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 int MOD = 1e9 + 7;
const int N = 2222;
int n, K;
long long sum[N], f[N][N];
char s[N];
long long work() {
  f[0][0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    int dt = n - i + 1;
    for (int j = 0; j <= K; j++) {
      for (int k = i - 1; k >= 0 && (i - k) * dt <= j; k--) {
        f[i][j] = (f[i][j] + f[k][j - (i - k) * dt]) % MOD;
      }
      f[i][j] = f[i][j] * ('z' - s[i]) % MOD;
    }
    for (int j = 0; j <= K; j++) {
      f[i][j] = (f[i][j] + sum[j] * (s[i] - 'a') % MOD) % MOD;
      sum[j] = (sum[j] + f[i][j]) % MOD;
    }
  }
  return sum[K];
}
int main() {
  scanf("%d%d%s", &n, &K, s + 1);
  printf("%lld\n", work());
  return 0;
}
