Levko loves strings of length n, consisting of lowercase English letters, very much. He has one such string s. For each string t of length n, Levko defines its beauty relative to s as the number of pairs of indexes i, j (1 ≤ i ≤ j ≤ n), such that substring t[i..j] is lexicographically larger than substring s[i..j].

The boy wondered how many strings t are there, such that their beauty relative to s equals exactly k. Help him, find the remainder after division this number by 1000000007 (109 + 7).

A substring s[i..j] of string s = s1s2... sn is string sisi + 1... sj.

String x = x1x2... xp is lexicographically larger than string y = y1y2... yp, if there is such number r (r < p), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1. The string characters are compared by their ASCII codes.

Input

The first line contains two integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 2000).

The second line contains a non-empty string s of length n. String s consists only of lowercase English letters. 

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

2 2
yz


Output

26


Input

2 3
yx


Output

2


Input

4 7
abcd


Output

21962
#include <bits/stdc++.h>
using namespace std;
const int N = 2005, P = 1000000007;
int n, m, f[N][N][2];
char s[N];
long long z;
void upd(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
int main() {
  scanf("%d%d%s", &n, &m, s), f[0][0][1] = 1;
  for (int i = 0; i < n + 1; ++i)
    for (int j = 0; j < m + 1; ++j)
      for (int t = 0; t < 2; ++t) {
        if (t && i)
          for (int k = 0; k < i && j + (n - i + 1) * (k + 1) <= m; ++k)
            upd(f[i][j + (n - i + 1) * (k + 1)][1],
                f[i - k - 1][j][1] * (122LL - s[i - 1]) % P);
        if (z = f[i][j][t])
          upd(f[i + 1][j][1], z * (s[i] - 97) % P), upd(f[i + 1][j][0], z);
      }
  printf("%d\n", (f[n][m][0] + f[n][m][1]) % P);
  return 0;
}
