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;
signed long long int dp[2005][2005];
signed long long int sdp[2005][2005];
int main() {
  int N, K;
  string S;
  cin >> N >> K >> S;
  S = " " + S;
  dp[0][0] = 1;
  sdp[0][0] = 1;
  for (int i = 1; i <= N; ++i) {
    for (int k = 0; k <= K; ++k) {
      dp[i][k] = sdp[i - 1][k] * (S[i] - 'a');
      dp[i][k] %= 1000000007;
      for (int j = i; j >= 1 and (i - j + 1) * (N - i + 1) <= k; --j) {
        dp[i][k] += dp[j - 1][k - (i - j + 1) * (N - i + 1)] * ('z' - S[i]);
        dp[i][k] %= 1000000007;
      }
      sdp[i][k] = (sdp[i - 1][k] + dp[i][k]) % 1000000007;
    }
  }
  cout << sdp[N][K];
}
