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 Sn = 2020;
const int mod = (int)1e9 + 7;
int n, k;
char s[Sn];
int a[Sn], U[Sn], D[Sn];
int dp[Sn][Sn];
int dpS[Sn];
int Ans;
void add(int& a, int b) { a = (a + b) % mod; }
int main() {
  scanf("%d%d%s", &n, &k, s + 1);
  for (int i = 1; i <= n; i++) a[i] = s[i] - 'a', U[i] = 25 - a[i], D[i] = a[i];
  dp[0][0] = 1;
  if (!k) add(Ans, 1);
  add(dpS[0], 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      add(dp[i][j], (long long)dpS[j] * D[i] % mod);
      for (int i1 = i - 1; i1 >= 0; i1--) {
        int ext = (i - i1) * (n - i + 1);
        if (j - ext < 0) break;
        add(dp[i][j], (long long)dp[i1][j - ext] * U[i] % mod);
      }
      add(dpS[j], dp[i][j]);
    }
    add(Ans, dp[i][k]);
  }
  cout << Ans << endl;
}
