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;
using ll = long long;
const int N = 2002;
const ll MOD = 1e9 + 7;
int n, k;
char s[N];
ll f[N][N][3];
inline void Add(ll& x, ll y) { (x += y) %= MOD; }
void Solve() {
  f[0][0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= k; ++j) {
      Add(f[i][j][1], f[i - 1][j][0] + f[i - 1][j][1] + f[i - 1][j][2]);
      Add(f[i][j][0], f[i][j][1] * (s[i - 1] - 'a'));
      for (int t = 0; t < j / (n - i + 1); ++t) {
        if (i >= t + 1) {
          int u = i - t - 1;
          ll tmp1 = f[u][j - (t + 1) * (n - i + 1)][0] +
                    f[u][j - (t + 1) * (n - i + 1)][2];
          ll tmp = tmp1 * ('z' - s[i - 1]);
          Add(f[i][j][2], tmp);
        }
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  scanf("%s", s);
  Solve();
  ll ans = f[n][k][0];
  Add(ans, f[n][k][1]);
  Add(ans, f[n][k][2]);
  printf("%lld\n", ans);
  return 0;
}
