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;
long long dpo[2003][2003];
long long dp[2003][2003];
long long sol(int p, int r, int n, string& s);
long long solo(int p, int r, int n, string& s) {
  if (p == n) return 0;
  if (dpo[p][r] >= 0) return dpo[p][r];
  long long& res = dpo[p][r] =
      ((s[p] - 'a') * sol(p + 1, r, n, s) + solo(p + 1, r, n, s)) % mod;
  return res;
}
long long sol(int p, int r, int n, string& s) {
  if (r < 0) return 0;
  if (p == n) return !r;
  if (dp[p][r] >= 0) return dp[p][r];
  long long& res = dp[p][r] = solo(p, r, n, s);
  if (r == 0) ++res;
  int e = 1, g = n - p;
  while (g and e <= g and r >= e * g) {
    long long big = ('z' - s[n - g]) * sol(n - g + 1, r - e * g, n, s);
    res = (res + big) % mod;
    if (e < g) {
      swap(e, g);
      long long big = ('z' - s[n - g]) * sol(n - g + 1, r - e * g, n, s);
      res = (res + big) % mod;
      swap(e, g);
    }
    ++e, --g;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  string s;
  while (cin >> n >> k >> s) {
    memset(dpo, -1, sizeof(dpo));
    memset(dp, -1, sizeof(dp));
    cout << sol(0, k, n, s) << '\n';
  }
  return 0;
}
