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 maxn = 2010;
const long long mod = 1000000007;
long long dp[maxn][maxn], sum[maxn];
char s[maxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  dp[0][0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      for (int l = i - 1; l >= 0 && j - (n - i + 1) * (i - l) >= 0; l--) {
        dp[i][j] = (dp[i][j] + ((long long)('z' - s[i]) *
                                dp[l][j - (n - i + 1) * (i - l)]) %
                                   mod) %
                   mod;
      }
      dp[i][j] = (dp[i][j] + ((long long)(s[i] - 'a') * sum[j]) % mod) % mod;
      sum[j] = (sum[j] + dp[i][j]) % mod;
    }
  }
  long long ans = 0;
  for (int i = 0; i <= n; i++) {
    ans = (ans + dp[i][k]) % mod;
  }
  printf("%lld\n", ans);
}
