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;
template <typename tn>
void read(tn &a) {
  tn x = 0, f = 1;
  char c = ' ';
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  a = x * f;
}
const int N = 101000, mod = 1e9 + 7;
int n, k;
char c[2010];
int f[2010][2010], s[2010];
void upd(int &a, long long b) { a = (a + b) % mod; }
int main() {
  read(n);
  read(k);
  cin >> c + 1;
  f[1][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) upd(s[j], f[i][j]);
    for (int j = 0; j <= k; j++) {
      upd(f[i + 1][j], 1ll * s[j] * (c[i] - 'a'));
      if (!f[i][j]) continue;
      int lst = 0;
      for (int p = 1; i + p <= n + 1 && p * (n - i - p + 2) + j <= k; p++) {
        lst = p;
        if (p * (n - i - p + 2) + j <= k)
          upd(f[i + p][j + p * (n - i - p + 2)],
              1ll * ('z' - c[i + p - 1]) * f[i][j]);
      }
      for (int p = n + 1 - i; p > lst && p * (n - i - p + 2) + j <= k; p--) {
        upd(f[i + p][j + p * (n - i - p + 2)],
            1ll * ('z' - c[i + p - 1]) * f[i][j]);
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i <= n + 1; i++) ans = (ans + f[i][k]) % mod;
  cout << ans << '\n';
  return 0;
}
