(Analysis by Richard Qi)

First, we show how to find the length of the shortest path between $a$ and $b$. It can be shown that greedily advancing from $a$ to the farthest right reachable interval is always optimal. So, from each interval, we can assign a unique "right parent" interval, and the distance from $a$ to $b$ can be found by repeatedly going to the parent of the current interval, ending when we reach an interval at $b$ or greater. To find these parents, we can use a two pointer technique where we scan from left to right and keep track of the parent of the current node, which always moves to the right. Naively, we can directly travel from $a$ to $b$ using these parent pointers.

To speed this process up to $\mathcal O(\log{N})$ time per query, we can use the technique of binary jumping. For every $k$ from $1 \dots \log{N}$, we can calculate the $2^k$-th parent of each interval: the $2^k$th parent of the $i$th interval is the $2^{k-1}$th parent of the $2^{k-1}$th parent of the $i$th interval. Then, to find the distance between $a$ and $b$, we can do the following process: for every $k$ from $\log{N} \dots 0$, we check whether the $2^k$th parent of $a$ is to the right of $b$ or not. If so, then the shortest path between $a$ and $b$ is less than $2^k$. Else, we recursively find the shortest path between this parent and $b$.

This observation nets us partial credit, as we can find the shortest path between $a$ and $b$, and then iterate over all special intervals $s$ between $a$ and $b$, and checking whether $dist(a, s)+dist(s, b) = dist(a, b)$. If so, then $s$ clearly lies on a shortest path from $a$ to $b$; otherwise, $dist(a, s) + dist(s, b) > dist(a, b)$ and any path going through $s$ must have distance greater than $dist(a, b)$.

An interesting observation to make is that when using the binary jumping solution above, we could arbitrarily use binary jumps going left or right. This will come in handy later on!

Now, we will take a closer look at the (not necessarily special) intervals $s$ which satisfy $dist(a, s) + dist(s, b) = dist(a, b)$. For convenience, let $L = dist(a, b)$. Consider the intervals $s$ which satisfy $dist(a, s) = i, dist(s, b) = L-i$ for some $i \in [1, L-1]$. From our greedy strategy for finding shortest paths mentioned above, it can be shown that the set of intervals which satisfy $dist(a, s) \le i$ are simply those with indices $\le greedy_a(i)$, where $greedy_a(i)$ denotes the interval we arrive at if we start at $a$ and greedily go right $i$ times. Similarly, the set of intervals which satisfy $dist(s, b) \le L-i$ are those with indices $\ge greedy_b(L-i)$, where $greedy_b(L-i)$ denotes the interval we arrive at if we start at $b$ and go greedily left $L-i$ times.

So, the intervals $s$ which satisfy $dist(a, s) = i, dist(s, b) = L-i$ are the intersection of these two sets: intervals which satisfy $greedy_b(L-i) \le s \le greedy_a(i)$, which forms some contiguous range of intervals. Notice that if we consider all of these ranges for all $i$, no interval is counted twice. Additionally, each range is nonempty (in particular, $greedy_a(i)$ always lies in range $i$). So, our answer can be expressed in the form $\sum_{i=1}^{L-1} (\text{number of special ranges between }greedy_b(L-i)\text{ and }greedy_a(i)\text{, inclusive})$.

(This potentially leaves out counting the intervals $a$ and $b$, which we can take care of as special cases.)

We can then use cumulative sums to simplify this expression. Define $csum(x)$ to be the number of special intervals between $1$ and $x$, inclusive. The sum can be rewritten as $\sum_{i=1}^{L-1} csum(greedy_a(i))-csum(greedy_b(L-i)-1)$. We can rewrite the sum as $\sum_{i=1}^{L-1} csum(greedy_a(i)) - \sum_{i=1}^{L-1} csum(greedy_b(L-i)-1) = \sum_{i=1}^{L-1} csum(greedy_a(i)) - \sum_{i=1}^{L-1} csum(greedy_b(i)-1)$.

Now, each of these sums can be computed independently and then summed up for the final answer. We show how to compute the second sum.

Consider again our binary jumping solution for computing the shortest path between $a$ and $b$, where we defined the parent of an interval $i$ as greedily going left one step. We can imagine these parent relations forming a graph (specifically, a forest) with directed edges going left. Then, the shortest path length is the length of the path starting at node $b$ and going up the tree until we arrive at a node with index $\le a$. For each node, define $par(node)$ to be the parent of the node. For each edge between some $node$ and $par(node)$, we can place a weight on the edge equal to $csum(par(node)-1)$. Then, the expression $\sum_{i=1}^{L-1} csum(greedy_b(i)-1)$ is simply the total weight of all the edges going from $b$ to its $L-1$th parent.

The sum of the weights of these edges can be computed again using binary jumps. For every $k$ from $1 \dots \log{n}$, we can precompute the sum of the weights on the path from each node $i$ and its $2^k$th parent. We do this recursively: the sum of the weights on the path from node $i$ and its $2^k$th parent is equal to the sum of the weights on the path from node $i$ to its $2^{k-1}$th parent plus the sum of the weights on the path from the $2^{k-1}$th parent of $i$ to the $2^{k-1}$th parent of the $2^{k-1}$th parent of $i$. We compute $\log{N}$ pieces of information for each of the nodes, which takes $\mathcal O(N \log{N})$ total time.

Then, to answer queries, we can traverse from $b$ to its $L-1$th parent using $\log{N}$ jumps (as described earlier), while summing up the weights we just computed every time we jump $2^k$ distance for some $k$.

#include <bits/stdc++.h>
using namespace std;
const int mx = 200005;
const int BIN = 30;
int N, Q;
//precompute binary jumps
vector<vector<pair<int, long long>>> genJmp(string S, vector<int> csum){
    vector<vector<pair<int, long long>>> res(N, vector<pair<int, long long>>(BIN));
    res[0][0] = make_pair(-1, 0);
    vector<pair<int, int>> rangs(N);
    int Ls = 0;
    int Rs = 0;
    for(int i = 0; i < int(S.size()); i++){
        if(S[i] == 'L'){
            rangs[Ls].first = i;
            rangs[Rs].second = i;
 	//calculate the information for the 2^0th parent (can also be done in linear time)
    set<pair<int, int>> rights;
    for(int i = 0; i < N; i++){
        if(i > 0){
            auto it = rights.lower_bound(make_pair(rangs[i].first, -1));
            int lower_ind = (it->second);
            assert(lower_ind < i);
            res[i][0] = make_pair(lower_ind, csum[lower_ind]);
        rights.insert(make_pair(rangs[i].second, i));
 	//generate binary jumps
    for(int i = 0; i < N; i++){
        for(int j = 1; j < BIN; j++){
            int lower_ind = res[i][j-1].first;
            assert(lower_ind < i);
            if(lower_ind == -1){
                res[i][j] = make_pair(-1, 0);
                res[i][j] = make_pair(res[lower_ind][j-1].first, res[i][j-1].second+res[lower_ind][j-1].second);
    return res;
int main() {
    cin >> N >> Q;
    string S; cin >> S;
    string special; cin >> special;

    //generate cumulative sums
    vector<int> csum(int(special.size()), 0);
    for(int i = 0; i < int(csum.size()); i++){
        if(i-1 >= 0) csum[i] = csum[i-1];
        csum[i]+=(special[i] == '1');

    vector<int> csum_left(N);
    //change indices of left cumulative sums
    for(int i = 1; i < N; i++){
        csum_left[i] = csum[i-1];
    //do a bunch of reversals so that we can use the same function for generating binary jumps, then reverse afterwards again
    vector<int> csum_right = csum; reverse(begin(csum_right), end(csum_right));
    vector<vector<pair<int, long long>>> jmp_left = genJmp(S, csum_left);

    reverse(begin(S), end(S));
    for(auto& u: S){
        u = ((u == 'L') ? 'R' : 'L');
    reverse(begin(special), end(special));

    vector<vector<pair<int, long long>>> jmp_right = genJmp(S, csum_right);

    reverse(begin(jmp_right), end(jmp_right));
    for(int i = 0; i < int(jmp_right.size()); i++){
        for(int j = 0; j < int(jmp_right[i].size()); j++){
            jmp_right[i][j].first = N-1-jmp_right[i][j].first;
    reverse(begin(S), end(S));
    for(auto& u: S){
        u = ((u == 'L') ? 'R' : 'L');
    reverse(begin(special), end(special));

    //answer queries
    for(int i = 1; i <= Q; i++){
        int a, b; cin >> a >> b;
        a--; b--; //0-indexed intervals

        int cur = b;
        long long sum = 0; //sum of edge weights going left from b
        int cur_right = a; //move this right as you go along
        long long sum_right = 0; //sum of edge weights going right from a
        int dist_traveled = 0;

        for(int j = BIN-1; j >= 0; j--){ //binary jump of size 2^j
            if(jmp_left[cur][j].first == -1) continue;
            if(jmp_left[cur][j].first > a){ //don't reach a or further to the left
                sum_right+=jmp_right[cur_right][j].second; //add up edge weights
                cur_right = jmp_right[cur_right][j].first; //update current node
                sum+=jmp_left[cur][j].second; //add up edge weights
                cur = jmp_left[cur][j].first; //update current node


                assert(cur_right != -1 && cur != -1);
        assert(cur_right < b);
        assert(cur > a);
        long long ans = sum_right-sum;

        if(special[a] == '1') ans++; //take care of a and b specially
        if(special[b] == '1') ans++;

        cout << dist_traveled+1 << " " << ans << "\n";
    // you should actually read the stuff at the bottom
/* stuff you should look for
    * int overflow, array bounds
    * special cases (n=1?)
    * do smth instead of nothing and stay organized

Bonus: Solve the case where the right endpoints of the intervals are not in increasing order (solution in this paper).