To generate the BST, FJ starts with a permutation $a=\{a_1,a_2,\ldots,a_N\}$ of the integers $1\ldots N$, where $N\le 300$. He then runs the following pseudocode with arguments $1$ and $N.$

generate(l,r):
  if l > r, return empty subtree;
  x = argmin_{l <= i <= r} a_i; // index of min a_i in {a_l,...,a_r}
  return a BST with x as the root, 
    generate(l,x-1) as the left subtree,
    generate(x+1,r) as the right subtree;

For example, the permutation $\{3,2,5,1,4\}$ generates the following BST:

    4
   / \
  2   5
 / \ 
1   3

Let $d_i(a)$ denote the depth of node $i$ in the tree corresponding to $a,$ meaning the number of nodes on the path from $a_i$ to the root. In the above example, $d_4(a)=1, d_2(a)=d_5(a)=2,$ and $d_1(a)=d_3(a)=3.$

The number of inversions of $a$ is equal to the number of pairs of integers $(i,j)$ such that $1\le i<j\le N$ and $a_i>a_j.$ The cows know that the $a$ that FJ will use to generate the BST has exactly $K$ inversions $(0\le K\le \frac{N(N-1)}{2})$. Over all $a$ satisfying this condition, compute the remainder when $\sum_ad_i(a)$ is divided by $M$ for each $1\le i\le N.$

SAMPLE INPUT:

3 0 192603497

SAMPLE OUTPUT:

1 2 3 

Here, the only permutation is $a=\{1,2,3\}.$

SAMPLE INPUT:

3 1 144408983

SAMPLE OUTPUT:

3 4 4 

Here, the two permutations are $a=\{1,3,2\}$ and $a=\{2,1,3\}.$

Problem credits: Yinzhan Xu