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>
long long cnt[2022][2022], total[2022], ans;
char s[2022];
int N, K, i, j, p, extra;
int main() {
  scanf("%d %d %s", &N, &K, s + 1);
  cnt[0][0] = total[0] = 1;
  ans = cnt[0][K];
  for (i = 1; i <= N; i++) {
    for (j = 0; j <= K; j++) {
      cnt[i][j] = (total[j] * (s[i] - 'a')) % 1000000007;
      for (extra = N - i + 1, p = i - 1; p >= 0 && extra <= j; p--) {
        cnt[i][j] = (cnt[i][j] + cnt[p][j - extra] * ('z' - s[i])) % 1000000007;
        extra += (N - i + 1);
      }
    }
    for (j = 0; j <= K; j++) total[j] = (total[j] + cnt[i][j]) % 1000000007;
    ans = (ans + cnt[i][K]) % 1000000007;
  }
  printf("%I64d\n", ans);
  return 0;
}
