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 dp[2005][2005];
int pre[2005][2005];
char s[2005];
int add(int x, int y) {
  return ((x + y) % 1000000007 + 1000000007) % 1000000007;
}
int mul(int x, int y) { return (long long)x * y % 1000000007; }
int mypow(int x, int c) {
  int ret = 1;
  while (c > 0) {
    if (c & 1) {
      ret = mul(ret, x);
    }
    c /= 2;
    x = mul(x, x);
  }
  return ret;
}
void solve() {
  int n, k;
  scanf("%d %d ", &n, &k);
  scanf("%s ", s + 1);
  dp[0][0] = 1;
  pre[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    int big = 'z' - s[i], small = s[i] - 'a';
    if (big > 0) {
      int w = n - i + 1;
      for (int t = 0; t <= k; ++t) {
        int g = 1;
        for (int last = i - 1; last >= 0; --last, ++g) {
          if (w * g + t <= k) {
            dp[i][w * g + t] = add(dp[i][w * g + t], mul(big, dp[last][t]));
          } else {
            break;
          }
        }
      }
    }
    if (small > 0) {
      for (int j = 0; j <= k; ++j) {
        dp[i][j] = add(dp[i][j], mul(small, pre[i - 1][j]));
      }
    }
    for (int j = 0; j <= k; ++j) {
      pre[i][j] = add(pre[i - 1][j], dp[i][j]);
    }
  }
  int ret = pre[n][k];
  printf("%d\n", ret);
}
int main() {
  solve();
  return 0;
}
