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 md = 1000000007;
int n, m, i, j, k, e, r, c[2020], o[2020], f[2020][2020];
char s[2020];
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s);
  for (i = 0; i < n; i++) c[i] = s[i] - 'a';
  f[0][0] = 1;
  for (i = 0; i <= n; i++)
    for (j = 0; j <= m; j++) {
      o[j] = (o[j] + f[i][j]) % md;
      if (i < n && s[i] > 'a')
        f[i + 1][j] = (f[i + 1][j] + ((long long)c[i]) * o[j]) % md;
      if (i < n && s[i] < 'z') {
        long long z = 25 - c[i];
        for (k = i, e = n - i; k >= 0 && j + e <= m; k--, e += n - i)
          f[i + 1][j + e] = (f[i + 1][j + e] + z * f[k][j]) % md;
      }
    }
  printf("%d\n", o[m]);
  return 0;
}
