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 = 2e3 + 7;
const long long md = 1e9 + 7;
long long dp[maxn][maxn];
long long pref[maxn];
int main() {
  int n, K;
  scanf("%d", &n);
  scanf("%d", &K);
  string s;
  cin >> s;
  s = " " + s;
  for (int j = 0; j < maxn; j++) {
    dp[0][j] = pref[j] = 0;
  }
  dp[0][0] = pref[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int k = 0; k <= K; k++) {
      dp[i][k] = pref[k] * ((int)(s[i] - 'a'));
      int cur = n - i + 1;
      for (int p = 1; (p * cur <= k) && (p <= i); p++) {
        dp[i][k] = dp[i][k] + dp[i - p][k - p * cur] * ((int)('z' - s[i]));
      }
      dp[i][k] %= md;
      pref[k] = (pref[k] + dp[i][k]) % md;
    }
  }
  printf("%lld", pref[K]);
  printf("\n");
}
