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 dp[2020][2020];
int now_lower[2020][2020];
char st[2020];
const int mod = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int multi(int a, int b) { return ((1ll * a * b) % mod); }
pair<int, int> calc(int a, int b) {
  int D = a * a - 4 * b;
  if (D <= 0) return make_pair(mod, mod);
  int c = (int)sqrt(1.0 * D);
  return make_pair((a - c) / 2 + 1, (a + c) / 2 - 1);
}
int main() {
  int n, k;
  cin >> n >> k >> st;
  dp[n][0] = 1;
  dp[n - 1][0] = st[n - 1] - 'a' + 1;
  dp[n - 1][1] = 'z' - st[n - 1];
  now_lower[n - 1][0] = st[n - 1] - 'a' + 1;
  for (int i = n - 2; i >= 0; i--) {
    for (int c = 0; c <= k; c++) {
      int temp = multi(dp[i + 1][c], st[i] - 'a');
      add(now_lower[i][c], now_lower[i + 1][c]);
      add(now_lower[i][c], temp);
      add(dp[i][c], now_lower[i][c]);
      if (c >= n - i) add(dp[i][c], multi(dp[i + 1][c - n + i], 'z' - st[i]));
      pair<int, int> ans = calc(n - i + 1, c);
      int until = min(n, i + ans.first);
      for (int itr = i + 1; itr < until; itr++) {
        int x = itr - i + 1;
        int y = n - itr;
        if (x * y > c) continue;
        add(dp[i][c], multi(dp[itr + 1][c - x * y], 'z' - st[itr]));
      }
      until = max(until, ans.second + i - 1);
      until = max(until, i + 1);
      for (int itr = until; itr < n; itr++) {
        int x = itr - i + 1;
        int y = n - itr;
        if (x * y > c) continue;
        add(dp[i][c], multi(dp[itr + 1][c - x * y], 'z' - st[itr]));
      }
    }
  }
  cout << dp[0][k] << endl;
  return 0;
}
