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>
int f[2005][2005], sum[2005][2005], n, m;
char s[2005];
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  sum[0][0] = f[0][0] = 1;
  for (int j = 0; j <= m; j++) {
    for (int i = 1; i <= n; i++) {
      int val = n - i + 1;
      for (int k = i - 1; k >= 0; k--)
        if (j - val >= 0) {
          f[i][j] = (f[i][j] + 1LL * f[k][j - val] * ('z' - s[i])) % 1000000007;
          val += n - i + 1;
        } else
          break;
      f[i][j] = (f[i][j] + 1LL * sum[i - 1][j] * (s[i] - 'a')) % 1000000007;
      sum[i][j] = (sum[i - 1][j] + f[i][j]) % 1000000007;
    }
  }
  printf("%d\n", (f[n][m] + sum[n - 1][m]) % 1000000007);
}
