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;
int N, M;
char S[2020];
int F[2020][2020], G[2020][2020], H[2020][2020];
int P[2020];
int main() {
  scanf("%d%d", &N, &M);
  scanf("%s", S + 1);
  memset(F, 0, sizeof(F));
  memset(G, 0, sizeof(G));
  memset(H, 0, sizeof(H));
  memset(P, 0, sizeof(P));
  H[N + 1][0] = 1;
  for (int i = N; i >= 1; i--) {
    for (int j = 0; j <= M; j++) {
      int s, t;
      t = j - (N - i + 1);
      if (t >= 0) {
        s = ((long long)F[i + 1][t] + G[i + 1][t] + H[i + 1][t]) % MOD;
        F[i][j] =
            ((long long)F[i][j] + (long long)s * ('z' - S[i]) % MOD) % MOD;
      }
      s = ((long long)F[i + 1][j] + G[i + 1][j] + H[i + 1][j]) % MOD;
      G[i][j] = ((long long)G[i][j] + (long long)s * (S[i] - 'a') % MOD) % MOD;
      H[i][j] = ((long long)H[i][j] + P[j]) % MOD;
      for (int k = i - 1; k > 0; k--) {
        t = (i - k) * (N - i + 1) + j;
        if (t <= M)
          H[k][t] = ((long long)H[k][t] + F[i][j]) % MOD;
        else
          break;
      }
      if (j == 0) H[i][j] = ((long long)H[i][j] + 1) % MOD;
    }
    for (int j = 0; j <= M; j++) P[j] = ((long long)P[j] + G[i][j]) % MOD;
  }
  int ans = ((long long)F[1][M] + G[1][M] + H[1][M]) % MOD;
  printf("%d\n", ans);
  return 0;
}
