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 N = 2001, INF = 1e9 + 7;
long long g[N][N], f[N][N];
int main() {
  int n, k;
  std::cin >> n >> k;
  std::string s;
  std::cin >> s;
  f[0][0] = g[0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= k; ++j) {
      g[i][j] = f[i - 1][j] * (s[n - i] - 'a') + g[i - 1][j];
      f[i][j] = g[i][j];
      for (int p = 1; p <= i + 1 - p && p * (i + 1 - p) <= j; ++p) {
        f[i][j] += f[i - p][j - p * (i + 1 - p)] * ('z' - s[n - i + p - 1]);
        if (p < i + 1 - p)
          f[i][j] += f[p - 1][j - p * (i + 1 - p)] * ('z' - s[n - p]);
      }
      g[i][j] %= INF;
      f[i][j] %= INF;
    }
  std::cout << f[n][k];
  return 0;
}
