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 = 1000000007;
int n, k, dp[2010][2010][2];
char a[2010];
vector<pair<int, int> > V[2010];
void update(int idx, int i, int kolko, int jednak, int val) {
  dp[i][kolko][jednak] += val;
  if (dp[i][kolko][jednak] >= MOD) dp[i][kolko][jednak] %= MOD;
}
int main() {
  scanf("%d%d%s", &n, &k, a);
  dp[0][0][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j + i < n; ++j)
      V[i].push_back(pair<int, int>((j + 1) * (n - i - j), j));
    sort(V[i].begin(), V[i].end());
  }
  for (int i = 0; i < n; ++i) {
    for (int J = 0; J <= k; ++J) {
      for (int ii = 0; ii < 2; ++ii) {
        if (dp[i][J][ii] == 0) continue;
        if (ii != 1) {
          for (int jj = 0; jj < (int)V[i].size(); ++jj) {
            int K = J + V[i][jj].first;
            int j = V[i][jj].second;
            if (K > k) break;
            update(i + j + 1, i + j + 1, K, 0,
                   ((long long)dp[i][J][ii] * (25 - (a[i + j] - 'a'))) % MOD);
          }
        }
        update(i + 1, i + 1, J, 1, dp[i][J][ii]);
        update(i + 1, i + 1, J, 0,
               ((long long)dp[i][J][ii] * (a[i] - 'a')) % MOD);
      }
    }
  }
  printf("%d\n", (dp[n][k][0] + dp[n][k][1]) % MOD);
  return 0;
}
