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;
long long dp[2005][2005];
long long dp2[2005][2005];
int n, k;
char s[2005];
int main() {
  srand((unsigned int)time(NULL));
  scanf("%d %d %s", &n, &k, &s);
  dp[n][0] = dp2[n][0] = 1;
  for (int i = n - 1; i >= 0; i--) {
    int bel = s[i] - 'a';
    for (int j = 0; j <= k; j++) {
      long long x = 1LL * bel * dp[i + 1][j] + dp2[i + 1][j];
      x %= 1000000007;
      dp[i][j] = dp2[i][j] = x;
    }
    for (int j = i; j < n; j++) {
      int a = (j - i + 1) * (n - j);
      for (int kk = a; kk <= k; kk++) {
        dp[i][kk] =
            (dp[i][kk] + 1LL * ('z' - s[j]) * dp[j + 1][kk - a]) % 1000000007;
      }
    }
  }
  printf("%lld\n", dp[0][k]);
}
