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;
int n, K, P = 7 + 1e9, sum[2010], f[2010][2010];
char s[2010];
void read(int &x) {
  char ch = getchar();
  int mark = 1;
  for (; ch != '-' && (ch < '0' || ch > '9'); ch = getchar())
    ;
  if (ch == '-') mark = -1, ch = getchar();
  for (x = 0; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
  x *= mark;
}
int main() {
  read(n);
  read(K);
  scanf("%s", s + 1);
  f[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= K; j++) (sum[j] += f[i - 1][j]) %= P;
    for (int j = 0; j <= K; j++)
      (f[i][j] += (long long)sum[j] * (s[i] - 'a') % P) %= P;
    for (int j = 0; j <= K; j++)
      for (int pre = i - 1; pre >= 0 && (n - i + 1) * (i - pre) + j <= K; pre--)
        (f[i][(n - i + 1) * (i - pre) + j] +=
         (long long)f[pre][j] * ('z' - s[i]) % P) %= P;
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) (ans += f[i][K]) %= P;
  printf("%d\n", ans);
  return 0;
}
