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 = 2020;
const int maxx = 10000;
const int MOd = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= MOd) a -= MOd;
}
int mul(int a, int b) { return (long long)a * b % MOd; }
int a, K, dn[maxn][maxn], top[maxn];
char ar[maxn];
int main() {
  scanf("%d %d", &a, &K);
  scanf(" %s", ar + 1);
  for (int i = 1; i <= a; i++) ar[i] = i[ar] - 'a';
  dn[0][0] = 1;
  top[0] = 1;
  for (int i = 1; i <= a; i++) {
    for (int j = 0; j <= K; j++) {
      if (ar[i] != 25) {
        int h = a - i + 1;
        for (int k = i - 1; k >= 0 && (i - k) * h <= j; k--) {
          add(dn[i][j], mul(dn[k][j - (i - k) * h], 25 - ar[i]));
        }
      }
      if (ar[i]) {
        add(dn[i][j], mul(top[j], ar[i]));
      }
      add(top[j], dn[i][j]);
    }
  }
  int ans = 0;
  for (int i = 0; i <= a; i++) add(ans, dn[i][K]);
  cout << ans << endl;
  return 0;
}
