We'll start by solving some simpler versions of this problem and using these to build to a full solution on the original problem. We'll assume familiarity with segment trees in this solution.

In the simplest version of this problem, imagine that the graph is a linked list and there are no modifications. We can compute prefix XORs along the linked list to answer any query in constant time.

If we add modifications to this version, we can maintain a segment tree on the linked list. An internal node maintains the XOR of all vertices that are inside the segment of the linked list that it covers. We can update the value at a given vertex in $O(\log N)$ time as a result. We'll come back to this idea of using a segment tree later.

Instead of solving this problem on a linked list, we'll work on solving it for an arbitrary tree but with no modifications. A subroutine that will be useful is efficiently finding the lowest common ancestor (LCA) of two arbitrary vertices. For each vertex, we can precompute the $2^d$th ancestor of that vertex - the $2^d$th ancestor for $d > 0$ is just the $2^{d-1}$th ancestor of the $2^{d-1}$th ancestor. To compute the LCA of two arbitrary vertices, we first go up the ancestor of the one lower in the tree until the two vertices are at the same height. We can then push both vertices up the tree as long as the $2^d$th ancestors differ, decreasing $d$ until it becomes zero. This runs in $O(\log N)$. We can augment this subroutine by, for each $2^d$th ancestor, adding the XOR of all values that we encounter going up the tree.

Adding in modifications is tricky - there are too many values to change if we explicitly try to maintain this data structure as accurately as possible. It is at this point that we will revisit trying to use a segment tree to store the XOR values. There are a few problems - primarily that a segment tree that backs an array guarantees that all the values that we want to XOR are contiguous, but on a tree the labeled vertices may not form a contiguous subarray.

We can mitigate this partially with a technique known as heavy-light decomposition. We'll start by labeling the edges of the graph as being either heavy or light. For a given vertex $v$ that is not a leaf, enumerate all of its children and the sizes of their subtrees. The child $c$ with the largest subtree has the edge $(v, c)$ get labeled as heavy. All other child edges out of $v$ are light.

We claim that any path between two vertices can go through at most $O(\log N)$ light edges. The reason for this is that in going up a light edge in the tree, the number of vertices in that tree must increase by more than a factor of 2. Therefore, every path going from any given vertex to the root cannot visit more than $O(\log N)$ light edges. The heavy edges are a different story. However, collections of adjacent heavy edges form long chains. The most extreme example of this is in the case of a linked list, in which case every edge in the graph is considered heavy.

After identifying whether each edge is heavy or light, we can relabel the vertices as follows - the root of the tree gets label 0. DFS down the tree, prioritizing heavy edges over light edges. Every time a new vertex is seen, assign it the next label that is available.

Note that the labels on a collection of heavy edges now form a contiguous subinterval, so now we can maintain a segment tree leveraging these new labels. We can then use this segment tree to compute the XOR of all values from some child to any ancestor. We check if the parent is connected via a heavy edge or a light edge. If it's a light edge, we travel it directly. Otherwise, we look at the top of the chain and go to the lower of the top of the chain and the ancestor.

#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 100000;
int segtree[4 * MAX_N];

void segtreeupd(int idx, int l, int r, int i, int v) {
  if(l == r) segtree[idx] = v;
  else {
    int m = (l+r)/2;
    if(i <= m) segtreeupd(2*idx, l, m, i, v);
    else segtreeupd(2*idx+1, m+1, r, i, v);
    segtree[idx] = segtree[2*idx] ^ segtree[2*idx+1];
  }
}
void segtreeupd(int i, int v) {
  segtreeupd(1, 0, MAX_N-1, i, v);
}

int segtreeqry(int idx, int l, int r, int lhs, int rhs) {
  if(l >= lhs && r <= rhs) return segtree[idx];
  int ret = 0;
  int m = (l+r)/2;
  if(m >= lhs) ret ^= segtreeqry(2*idx, l, m, lhs, rhs);
  if(m+1 <= rhs) ret ^= segtreeqry(2*idx+1, m+1, r, lhs, rhs);
  return ret;
}
int segtreeqry(int l, int r) {
  return segtreeqry(1, 0, MAX_N-1, l, r);
}

const int MAX_D = 17;
int lca[MAX_N][MAX_D];
int depth[MAX_N];

int getLCA(int a, int b) {
  if(depth[a] < depth[b]) swap(a, b);
  for(int d = MAX_D-1; d >= 0; d--) {
    if(depth[a] - (1<<d) >= depth[b]) {
      a = lca[a][d];
    }
  }
  for(int d = MAX_D-1; d >= 0; d--) {
    if(lca[a][d] != lca[b][d]) {
      a = lca[a][d];
      b = lca[b][d];
    }
  }
  if(a != b) {
    a = lca[a][0];
    b = lca[b][0];
  }
  return a;
}

void initLCA() {
  for(int d = 1; d < MAX_D; d++) {
    for(int i = 0; i < MAX_N; i++) {
      lca[i][d] = lca[lca[i][d-1]][d-1];
    }
  }
}

vector<int> edges[MAX_N];
int treesz[MAX_N];
int vertextosegtree[MAX_N];
int topchain[MAX_N];
int vals[MAX_N];

void dfsForHLD(int curr, int topPtr, int par, int& internalsegtreeidx) {
  vertextosegtree[curr] = internalsegtreeidx++;
  segtreeupd(vertextosegtree[curr], vals[curr]);
  topchain[curr] = topPtr;
  int largestchild = -1;
  int largestsz = -1;
  for(int out: edges[curr]) {
    if(out == par) continue;
    if(treesz[out] > largestsz) {
      largestsz = treesz[out];
      largestchild = out;
    }
  }
  if(largestchild < 0) return;
  dfsForHLD(largestchild, topPtr, curr, internalsegtreeidx);
  for(int out: edges[curr]) {
    if(out == par || out == largestchild) continue;
    dfsForHLD(out, out, curr, internalsegtreeidx);
  }
}

void dfsForSize(int curr, int par) {
  treesz[curr]++;
  for(int out: edges[curr]) {
    if(out == par) continue;
    depth[out] = depth[curr] + 1;
    lca[out][0] = curr;
    dfsForSize(out, curr);
    treesz[curr] += treesz[out];
  }
}

void initHLD() {
  dfsForSize(0, -1);
  initLCA();
  int internalsegtreeidx = 0;
  dfsForHLD(0, 0, -1, internalsegtreeidx);
}

int pathQuery(int child, int par) {
  int ret = 0;
  while(child != par) {
    if(topchain[child] == child) {
      // light edge
      ret ^= vals[child];
      child = lca[child][0];
    }
    else if(depth[topchain[child]] > depth[par]) {
      ret ^= segtreeqry(vertextosegtree[topchain[child]], vertextosegtree[child]);
      child = lca[topchain[child]][0];
    }
    else {
      ret ^= segtreeqry(vertextosegtree[par]+1, vertextosegtree[child]);
      break;
    }
  }
  return ret;
}

int query(int a, int b) {
  int r = getLCA(a, b);
  return pathQuery(a, r) ^ pathQuery(b, r) ^ vals[r];
}

int main() {
  freopen("cowland.in", "r", stdin);
  freopen("cowland.out", "w", stdout);
  int n, q;
  cin >> n >> q;
  for(int i = 0; i < n; i++) {
    cin >> vals[i];
  }
  for(int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--; b--;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  initHLD();
  while(q--) {
    int t;
    cin >> t;
    if(t == 1) {
      int i, v;
      cin >> i >> v;
      vals[--i] = v;
      segtreeupd(vertextosegtree[i], v);
    }
    else {
      int a, b;
      cin >> a >> b;
      cout << query(--a, --b) << "\n";
    }
  }
}