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;
char s[2005];
int dp[2005][2005], dp2[2005][2005];
int sum[2005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  scanf("%s", s + 1);
  sum[n + 1] = 1;
  for (int i = n; i >= 1; i--) {
    int x = (s[i] - 'a' + 1);
    sum[i] = (sum[i + 1] * 1ll * x) % mod;
  }
  if (k == 0) {
    printf("%d\n", sum[1]);
    return 0;
  }
  dp[0][0] = 1;
  dp2[0][0] = 1;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int cnt = 25 - (s[i] - 'a');
    for (int x = 0; x <= k; x++) {
      dp[i][x] = (dp[i][x] + dp2[i - 1][x] * 1ll * (s[i] - 'a')) % mod;
    }
    for (int j = i; j >= 1; j--) {
      int l = i - j + 1, r = n - i + 1;
      if (l * r > k) break;
      for (int x = 0; x <= k; x++) {
        if (dp[j - 1][x] == 0) continue;
        if (x + l * r <= k)
          dp[i][x + l * r] =
              (dp[i][x + l * r] + dp[j - 1][x] * 1ll * cnt % mod) % mod;
        if (x + l * r == k)
          ans = (ans + dp[j - 1][x] * 1ll * cnt % mod * sum[i + 1] % mod) % mod;
      }
    }
    for (int x = 0; x <= k; x++) {
      dp2[i][x] = (dp2[i - 1][x] + dp[i][x]) % mod;
    }
  }
  printf("%lld\n", ans);
}
