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;
string s;
int f[2222][2222][2];
const int mod = 1000000007;
int sum[2222][2222];
inline void add(int i, int j, int k, int val) {
  f[i][j][k] += val;
  if (f[i][j][k] >= mod) f[i][j][k] -= mod;
}
int solve() {
  int ans = 0;
  for (int pos = 0; pos < n; ++pos) {
    int ls = s[pos] - 'a';
    int mr = 25 - ls;
    for (int cnt = 0; cnt <= k; ++cnt) {
      add(pos, cnt, 0, ((long long)ls * (sum[cnt][pos - 1])) % mod);
      if (!cnt) {
        add(pos, cnt, 0, ls);
      }
      for (int prev = pos - 1; prev >= 0; --prev) {
        int to_add = (pos - prev) * (n - pos);
        if (to_add > cnt) break;
        add(pos, cnt, 1, ((long long)mr * f[prev][cnt - to_add][0]) % mod);
        add(pos, cnt, 1, ((long long)mr * f[prev][cnt - to_add][1]) % mod);
      }
      int to_add = (pos + 1) * (n - pos);
      if (to_add == cnt) {
        add(pos, cnt, 1, mr);
      }
      sum[cnt][pos] += f[pos][cnt][0];
      if (sum[cnt][pos] >= mod) sum[cnt][pos] -= mod;
      sum[cnt][pos] += f[pos][cnt][1];
      if (sum[cnt][pos] >= mod) sum[cnt][pos] -= mod;
      sum[cnt][pos] += sum[cnt][pos - 1];
      if (sum[cnt][pos] >= mod) sum[cnt][pos] -= mod;
    }
    ans += f[pos][k][0];
    if (ans >= mod) ans -= mod;
    ans += f[pos][k][1];
    if (ans >= mod) ans -= mod;
  }
  if (k == 0) {
    ++ans;
    if (ans == mod) ans = 0;
  }
  return ans;
}
int main() {
  cin >> n >> k;
  cin >> s;
  cout << solve() << endl;
}
