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;
char S[2005];
int dp[2005][2005];
int sum[2005][2005];
int main() {
  int N, K;
  cin >> N >> K;
  scanf("%s", S);
  dp[N][0] = 1;
  for (int i = (N); i >= (0); --i) {
    for (int k = (0); k <= (K); ++k) {
      if (i < N)
        dp[i][k] = (1ll * dp[i][k] + sum[i][k] + dp[N][k]) % 1000000007;
      if (!dp[i][k]) continue;
      if (i > 0)
        sum[i - 1][k] =
            (sum[i][k] + 1ll * (S[i - 1] - 'a') * dp[i][k]) % 1000000007;
      for (int n = (i - 1); n >= (0); --n) {
        int k2 = k + (i - n) * (N - i + 1);
        if (k2 > K) break;
        dp[n][k2] =
            (dp[n][k2] + 1ll * ('z' - S[i - 1]) * dp[i][k]) % 1000000007;
      }
    }
  }
  cout << dp[0][K] << endl;
}
