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>
const int maxn = 2001, maxm = 2001, mod = 1000000007;
int n, m, f[maxn][maxm];
char s[maxn];
void inc(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  f[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    s[i] -= 'a' - 1;
    for (int j = 0; j <= m; ++j) {
      f[i][j] = (long long)f[i - 1][j] * s[i] % mod;
      for (int k = 1; k <= i && k * (n - i + 1) <= j; ++k)
        inc(f[i][j],
            (long long)(f[i - k][j - k * (n - i + 1)] -
                        (k < i ? f[i - k - 1][j - k * (n - i + 1)] : 0) + mod) *
                (26 - s[i]) % mod);
    }
  }
  printf("%d\n", f[n][m]);
  return 0;
}
