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 inf = 1 << 29;
char str[2010];
long long dp[2010][2010][3];
int sum[2010];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  scanf("%s", str);
  memset(dp, 0, sizeof(dp));
  dp[n][0][0] = 1;
  for (int i = n - 1; i >= 0; i--) {
    dp[i][0][0] = (dp[i][0][0] + 1) % 1000000007;
    for (int j = 0; j <= k; j++) {
      for (int p = 0; p < 3; p++) {
        int t = str[i] - 'a';
        dp[i][j][1] += dp[i + 1][j][p] * t;
        dp[i][j][1] %= 1000000007;
        t = 'z' - str[i];
        if (j - (n - i) >= 0) {
          dp[i][j][2] += dp[i + 1][j - (n - i)][p] * t;
          dp[i][j][2] %= 1000000007;
        }
      }
      dp[i][j][0] += sum[j];
      dp[i][j][0] %= 1000000007;
    }
    for (int j = 0; j <= k; j++) {
      sum[j] += dp[i][j][1];
      sum[j] %= 1000000007;
      int t = j;
      for (int p = i - 1; p >= 0; p--) {
        t += n - i;
        if (t > k) break;
        dp[p][t][0] += dp[i][j][2];
        dp[p][t][0] %= 1000000007;
      }
    }
  }
  long long ans = (dp[0][k][0] + dp[0][k][1] + dp[0][k][2]) % 1000000007;
  printf("%I64d\n", ans);
}
