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[2003];
int dp[2003][2003][2];
int main() {
  int i, j, t;
  scanf("%d%d%s", &n, &m, s + 1);
  for (i = 1; i <= n; i++) s[i] ^= 0x60;
  dp[0][0][0] = 1;
  dp[0][0][1] = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= m; j++) {
      dp[i][j][1] = (dp[i - 1][j][0] + dp[i - 1][j][1]) % mod;
      dp[i][j][0] = (long long)(dp[i - 1][j][0] + dp[i - 1][j][1]) % mod *
                    (s[i] - 1) % mod;
      int r = n - i + 1;
      for (t = i - 1; j - (i - t) * r >= 0 && t >= 0; t--) {
        dp[i][j][0] += (long long)dp[t][j - (i - t) * r][0] * (26 - s[i]) % mod;
        dp[i][j][0] %= mod;
      }
    }
  }
  printf("%d\n", (dp[n][m][0] + dp[n][m][1]) % mod);
  return 0;
}
