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;
int n, k, F[2010][2010], G[2010][2010];
char s[2010];
inline void Add(int &x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
  if (x < 0) x += 1000000007;
}
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  G[0][0] = F[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      Add(F[i][j], 1ll * G[i - 1][j] * (s[i] - 'a') % 1000000007);
      for (int t = i - 1; j - (i - t) * (n - i + 1) >= 0 && t >= 0; t--)
        Add(F[i][j], 1ll * F[t][j - (i - t) * (n - i + 1)] *
                         (26 - s[i] + 'a' - 1) % 1000000007);
      G[i][j] = (G[i - 1][j] + F[i][j]) % 1000000007;
    }
  }
  cout << G[n][k] << endl;
  return 0;
}
