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 mod = int(1e9) + 7;
long long d[2020][2020], sum[2020], ans;
int n, K, Next;
string s;
int main() {
  cin >> n >> K >> s;
  d[0][0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int k = 0; k <= K; k++) {
      (d[i][k] += sum[k] * ((i == n) + s[i - 1] - 'a')) %= mod;
      for (int j = i; j >= 1; j--) {
        Next = (i - j + 1) * (n - i + 1);
        if (k < Next) break;
        (d[i][k] += d[j - 1][k - Next] * (26 - (s[i - 1] - 'a' + 1))) %= mod;
      }
    }
    for (int k = 0; k <= K; k++) (sum[k] += d[i][k]) %= mod;
  }
  cout << d[n][K];
  return 0;
}
