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;
const int maxn = 2020;
int dp[maxn][maxn], sum[maxn][maxn];
char s[maxn];
int n, k;
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%d %d %s", &n, &k, s + 1);
  dp[0][0] = 1;
  sum[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    int c = s[i] - 'a';
    for (int j = 0; j <= k; j++) {
      add(dp[i][j], 1ll * c * sum[j][i - 1] % mod);
      sum[j][i] = sum[j][i - 1];
      int now = 0;
      for (int t = i - 1; t >= 0; t--) {
        now += (n - i + 1);
        if (j + now > k) break;
        add(dp[i][j + now], 1ll * (25 - c) * dp[t][j] % mod);
      }
      add(sum[j][i], dp[i][j]);
    }
  }
  int res = 0;
  for (int i = 0; i <= n; i++) {
    add(res, dp[i][k]);
  }
  printf("%d\n", res);
  return 0;
}
