Problem: You are given a binary string s (each character of this string is either 0 or 1).

Let's denote the cost of string t as the number of occurences of s in t. For example, if s is 11 and t is 111011, then the cost of t is 3.

Let's also denote the Fibonacci strings sequence as follows:

  * F(0) is 0;
  * F(1) is 1;
  * F(i) = F(i - 1) + F(i - 2) if i > 1, where  +  means the concatenation of two strings.



Your task is to calculate the sum of costs of all subsequences of the string F(x). Since answer may be large, calculate it modulo 109 + 7.

Input

The first line contains two integers n and x (1 ≤ n ≤ 100, 0 ≤ x ≤ 100) — the length of s and the index of a Fibonacci string you are interested in, respectively.

The second line contains s — a string consisting of n characters. Each of these characters is either 0 or 1.

Output

Print the only integer — the sum of costs of all subsequences of the string F(x), taken modulo 109 + 7. 

Examples

Input

2 4
11


Output

14


Input

10 100
1010101010


Output

553403224

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int dp[105][105][105];
long long int dp2[105][105][105];
long long int dp3[105][105][105];
long long int dp4[105][105][105];
long long int modpow(long long int a, long long int b) {
  if (b == 0) {
    return 1;
  }
  long long int ans = modpow(a, b / 2);
  ans = (ans * ans) % 1000000007;
  if (b % 2) {
    ans = (ans * a) % 1000000007;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, x;
  cin >> n >> x;
  string s;
  cin >> s;
  for (int a = 1; a <= n; a++) {
    if (s.at(a - 1) == '0') {
      dp[0][a][a] = dp2[0][a][a] = dp3[0][a][a] = dp4[0][a][a] = 1;
    } else {
      dp[1][a][a] = dp2[1][a][a] = dp3[1][a][a] = dp4[1][a][a] = 1;
    }
  }
  long long int len = 1;
  long long int prevlen = 1;
  long long int prevnum = 2;
  long long int prevprevnum = 2;
  for (int i = 2; i <= x; i++) {
    long long int temp = len;
    len = (prevlen + len) % (1000000007 - 1);
    long long int num = modpow(2, len);
    prevlen = temp;
    for (int a = 1; a <= n; a++) {
      for (int b = a; b <= n; b++) {
        dp[i][a][b] += (prevnum * dp[i - 2][a][b]) % 1000000007;
        dp2[i][a][b] += (prevnum * dp2[i - 2][a][b]) % 1000000007;
        dp3[i][a][b] += dp3[i - 2][a][b];
        dp4[i][a][b] += dp4[i - 2][a][b];
        for (int infirst = 1; infirst < b - a + 1; infirst++) {
          dp[i][a][b] +=
              (dp2[i - 1][a][a + infirst - 1] * dp3[i - 2][a + infirst][b]) %
              1000000007;
          dp2[i][a][b] +=
              (dp2[i - 1][a][a + infirst - 1] * dp4[i - 2][a + infirst][b]) %
              1000000007;
          dp3[i][a][b] +=
              (dp4[i - 1][a][a + infirst - 1] * dp3[i - 2][a + infirst][b]) %
              1000000007;
          dp4[i][a][b] +=
              (dp4[i - 1][a][a + infirst - 1] * dp4[i - 2][a + infirst][b]) %
              1000000007;
        }
        dp[i][a][b] += (dp[i - 1][a][b] * prevprevnum) % 1000000007;
        dp2[i][a][b] += dp2[i - 1][a][b];
        dp3[i][a][b] += (dp3[i - 1][a][b] * prevprevnum) % 1000000007;
        dp4[i][a][b] += dp4[i - 1][a][b];
        dp[i][a][b] = dp[i][a][b] % 1000000007;
        dp2[i][a][b] = dp2[i][a][b] % 1000000007;
        dp3[i][a][b] = dp3[i][a][b] % 1000000007;
        dp4[i][a][b] = dp4[i][a][b] % 1000000007;
      }
    }
    prevprevnum = prevnum;
    prevnum = num;
  }
  cout << dp[x][1][n];
  return 0;
}