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;
int n, C;
long long int dp[2005][2005], sum[2005];
char s[2005];
int main() {
  scanf("%d%d", &n, &C);
  scanf("%s", s);
  memset(dp, 0, sizeof(dp));
  memset(sum, 0, sizeof(sum));
  sum[0] = dp[0][0] = 1;
  for (int i = (1); i <= (n); i++) {
    for (int j = (0); j <= (C); j++) {
      dp[i][j] = sum[j] * (s[i - 1] - 'a');
      int x = max(i - j / (n - i + 1), 0);
      for (int k = (x); k < (i); k++)
        dp[i][j] += dp[k][j - (i - k) * (n - i + 1)] * ('z' - s[i - 1]);
      dp[i][j] %= 1000000007;
      sum[j] = (sum[j] + dp[i][j]) % 1000000007;
    }
  }
  long long int ret = (C == 0) ? 1 : 0;
  for (int i = (1); i <= (n); i++) ret = (ret + dp[i][C]) % 1000000007;
  printf("%I64d\n", ret);
  return 0;
}
