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 P = 1e9 + 7;
int n, K;
char S[2005];
int dp[2005][2005], sum[2005];
void add(int &a, int b) {
  a += b;
  if (a >= P) a -= P;
}
int main() {
  scanf("%d %d %s", &n, &K, S + 1);
  dp[0][0] = 1, sum[0] = 1;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= K; j++) {
      for (int k = 0; k < i && (k + 1) * (n - i + 1) <= j; k++) {
        add(dp[i][j],
            1ll * dp[i - k - 1][j - (k + 1) * (n - i + 1)] * ('z' - S[i]) % P);
      }
      add(dp[i][j], 1ll * sum[j] * (S[i] - 'a') % P);
      add(sum[j], dp[i][j]);
    }
  }
  for (int i = 0; i <= n; i++) {
    add(ans, dp[i][K]);
  }
  printf("%d\n", ans);
  return 0;
}
