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>
char x[2222];
long long dp[2222][2222], sp[2222][2222];
int main() {
  int n, k;
  scanf("%d%d\n", &n, &k);
  gets(x + 1);
  dp[n + 1][0] = 1;
  for (int i = n; i >= 1; --i)
    for (int j = 0; j <= k; ++j) {
      int myK;
      for (int k = i; k <= n; ++k) {
        myK = k;
        int val = (k - i + 1) * (n - k + 1);
        if (val > j) break;
        dp[i][j] += dp[k + 1][j - val] * ('z' - x[k]);
        dp[i][j] %= 1000000007;
      }
      for (int k = n; k >= i && k > myK; --k) {
        int val = (k - i + 1) * (n - k + 1);
        if (val > j) break;
        dp[i][j] += dp[k + 1][j - val] * ('z' - x[k]);
        dp[i][j] %= 1000000007;
      }
      sp[i][j] = sp[i + 1][j] + dp[i + 1][j] * (x[i] - 'a');
      sp[i][j] %= 1000000007;
      dp[i][j] += sp[i][j];
      dp[i][j] %= 1000000007;
      if (j == 0) dp[i][j] = (dp[i][j] + 1) % 1000000007;
    }
  printf("%d", dp[1][k]);
  return 0;
}
