#include <bits/stdc++.h>
const int N = 2005;
const int mod = 1e9 + 7;
using namespace std;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int n, k, dp[N][N], sum[N][N];
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> s;
  dp[0][0] = sum[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      add(dp[i][j], 1LL * sum[i - 1][j] * (s[i - 1] - 'a') % mod);
      for (int h = i - 1; h >= 0; --h) {
        if (j - (i - h) * (n - i + 1) < 0) break;
        add(dp[i][j],
            1LL * dp[h][j - (i - h) * (n - i + 1)] * ('z' - s[i - 1]) % mod);
      }
      sum[i][j] = (sum[i - 1][j] + dp[i][j]) % mod;
    }
  }
  cout << sum[n][k];
}

### Prompt

Please create a solution in cpp to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
const int N = 2005;
const int mod = 1e9 + 7;
using namespace std;
void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int n, k, dp[N][N], sum[N][N];
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> s;
  dp[0][0] = sum[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      add(dp[i][j], 1LL * sum[i - 1][j] * (s[i - 1] - 'a') % mod);
      for (int h = i - 1; h >= 0; --h) {
        if (j - (i - h) * (n - i + 1) < 0) break;
        add(dp[i][j],
            1LL * dp[h][j - (i - h) * (n - i + 1)] * ('z' - s[i - 1]) % mod);
      }
      sum[i][j] = (sum[i - 1][j] + dp[i][j]) % mod;
    }
  }
  cout << sum[n][k];
}

```