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 = 2005;
const ll mod = 1e9 + 7;
ll add(ll a, ll b) { return (a + b) % mod; }
ll sub(ll a, ll b) { return (a - b + mod * mod) % mod; }
ll mul(ll a, ll b) { return (a * b) % mod; }
int n, k;
string s;
ll dp[N][N], sum[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> s;
  s = " " + s;
  dp[0][0] = sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      dp[i][j] = add(dp[i][j], mul(sum[j], s[i] - 'a'));
      ll res = 0;
      for (int l = i - 1; l >= 0; l--) {
        if ((i - l) * (n - i + 1) > j) break;
        res = add(res, dp[l][j - (i - l) * (n - i + 1)]);
      }
      dp[i][j] = add(dp[i][j], mul(res, 'z' - s[i]));
      sum[j] = add(sum[j], dp[i][j]);
    }
  }
  ll ans = 0;
  for (int i = 0; i <= n; i++) ans = add(ans, dp[i][k]);
  cout << ans;
}
