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;
inline void setmin(int &x, int y) {
  if (y < x) x = y;
}
inline void setmax(int &x, int y) {
  if (y > x) x = y;
}
inline void setmin(long long &x, long long y) {
  if (y < x) x = y;
}
inline void setmax(long long &x, long long y) {
  if (y > x) x = y;
}
const int N = 100;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
inline void add(long long &x, long long y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
long long pw[N + 1];
long long dp[N + 1][N][N][4];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int n, x;
  cin >> n >> x;
  pw[0] = pw[1] = 2;
  for (int i = 2; i < x + 1; i++) pw[i] = pw[i - 1] * pw[i - 2] % MOD;
  string s;
  cin >> s;
  for (int i = 0; i < n; i++)
    if (s[i] == '0') {
      for (int mask = 0; mask < 4; mask++) dp[0][i][i][mask] = 1;
    } else {
      for (int mask = 0; mask < 4; mask++) dp[1][i][i][mask] = 1;
    }
  for (int t = 2; t < x + 1; t++)
    for (int i = 0; i < n; i++)
      for (int j = i; j < n; j++) {
        add(dp[t][i][j][0], dp[t - 1][i][j][0]);
        add(dp[t][i][j][0], dp[t - 2][i][j][0]);
        for (int cnt = 1; cnt < j - i + 1; cnt++)
          add(dp[t][i][j][0],
              dp[t - 1][i][i + cnt - 1][0] * dp[t - 2][i + cnt][j][0] % MOD);
        add(dp[t][i][j][1], dp[t - 1][i][j][1]);
        add(dp[t][i][j][1], dp[t - 2][i][j][1] * pw[t - 1] % MOD);
        for (int cnt = 1; cnt < j - i + 1; cnt++)
          add(dp[t][i][j][1],
              dp[t - 1][i][i + cnt - 1][1] * dp[t - 2][i + cnt][j][0] % MOD);
        add(dp[t][i][j][2], dp[t - 1][i][j][2] * pw[t - 2] % MOD);
        add(dp[t][i][j][2], dp[t - 2][i][j][2]);
        for (int cnt = 1; cnt < j - i + 1; cnt++)
          add(dp[t][i][j][2],
              dp[t - 1][i][i + cnt - 1][0] * dp[t - 2][i + cnt][j][2] % MOD);
        add(dp[t][i][j][3], dp[t - 1][i][j][3] * pw[t - 2] % MOD);
        add(dp[t][i][j][3], dp[t - 2][i][j][3] * pw[t - 1] % MOD);
        for (int cnt = 1; cnt < j - i + 1; cnt++)
          add(dp[t][i][j][3],
              dp[t - 1][i][i + cnt - 1][1] * dp[t - 2][i + cnt][j][2] % MOD);
      }
  cout << dp[x][0][n - 1][3] << "\n";
  return 0;
}