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 maxn = 3000;
const long long P = 1000000000 + 7;
long long dp[maxn][maxn][2];
int main() {
  int N, K;
  string s;
  cin >> N >> K >> s;
  dp[0][0][0] = 1;
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j <= K; ++j) {
      dp[i + 1][j][0] += (dp[i][j][0] + dp[i][j][1]) * (s[i] - 'a');
      dp[i + 1][j][0] %= P;
      dp[i + 1][j][1] += (dp[i][j][0] + dp[i][j][1]);
      dp[i + 1][j][1] %= P;
      int k;
      for (k = 0; k + i < N && j + (k + 1) * (N - k - i) <= K; ++k) {
        dp[i + k + 1][j + (k + 1) * (N - k - i)][0] +=
            dp[i][j][0] * ('z' - s[i + k]);
        dp[i + k + 1][j + (k + 1) * (N - k - i)][0] %= P;
      }
      if (k + i < N) {
        for (k = N - i - 1; k >= 0 && j + (k + 1) * (N - k - i) <= K; --k) {
          dp[i + k + 1][j + (k + 1) * (N - k - i)][0] +=
              dp[i][j][0] * ('z' - s[i + k]);
          dp[i + k + 1][j + (k + 1) * (N - k - i)][0] %= P;
        }
      }
    }
  }
  long long ans = (dp[N][K][0] + dp[N][K][1]) % P;
  cout << ans << '\n';
  return 0;
}
