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 = 2005;
const int inf = 1 << 29;
const int mod = 1000000007;
const double eps = 1e-8;
char s[maxn];
int n, li, dp[maxn][maxn], sum[maxn];
int main() {
  scanf("%d%d%s", &n, &li, s + 1);
  dp[0][0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    long long tp = s[i] - 'a';
    for (int j = 0; j < n + 1 - i; j++) {
      dp[i][j] = tp * sum[j] % mod;
      sum[j] = (sum[j] + dp[i][j]) % mod;
    }
    for (int j = n + 1 - i; j <= li; j++) {
      for (int k = 1; k * (n + 1 - i) <= j && k <= i; k++)
        dp[i][j] = (dp[i][j] + dp[i - k][j - k * (n + 1 - i)]) % mod;
      dp[i][j] = (25 - tp) * dp[i][j] % mod;
      dp[i][j] = (dp[i][j] + tp * sum[j]) % mod;
      sum[j] = (sum[j] + dp[i][j]) % mod;
    }
  }
  printf("%d\n", sum[li]);
}
