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;
const long long MOD = 1e9 + 7;
long long n, k;
char ch[2010];
long long f[2010][2010];
int main() {
  while (~scanf("%I64d %I64d", &n, &k)) {
    scanf("%s", ch);
    memset(f, 0, sizeof f);
    f[0][0] = 1;
    for (long long i = 1; i <= n; i++) {
      for (long long j = 0; j <= k; j++) {
        f[i][j] = f[i - 1][j] * (long long)(ch[i - 1] - 'a' + 1) % MOD;
        for (long long m = i - 1; m >= 0 && (i - m) * (n - i + 1) <= j; m--) {
          long long tmp = j - (i - m) * (n - i + 1);
          if (m == 0)
            f[i][j] += f[m][tmp] * ('z' - ch[i - 1]) % MOD;
          else
            f[i][j] += (f[m][tmp] - f[m - 1][tmp]) * ('z' - ch[i - 1]) % MOD;
          f[i][j] %= MOD;
        }
      }
    }
    long long ans = (f[n][k] % MOD + MOD) % MOD;
    printf("%I64d\n", ans);
  }
  return 0;
}
