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 INF = 3e18;
const long long fact_table = 800008;
double Pi = 3.1415926535897932384626;
vector<pair<long long, long long> > tree[500010];
priority_queue<long long> pql;
priority_queue<pair<long long, long long> > pqp;
priority_queue<long long, vector<long long>, greater<long long> > pqls;
priority_queue<pair<long long, long long>, vector<pair<long long, long long> >,
               greater<pair<long long, long long> > >
    pqps;
long long dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
long long dy[8] = {0, 1, 0, -1, 1, -1, -1, 1};
char dir[] = "DRUL";
long long mod = 1000000007;
long long rui(long long number1, long long number2) {
  if (number2 == 0) {
    return 1;
  } else {
    long long number3 = rui(number1, number2 / 2);
    number3 *= number3;
    number3 %= mod;
    if (number2 % 2 == 1) {
      number3 *= number1;
      number3 %= mod;
    }
    return number3;
  }
}
long long gcd(long long number1, long long number2) {
  if (number1 > number2) {
    swap(number1, number2);
  }
  if (number1 == 0 || number1 == number2) {
    return number2;
  } else {
    return gcd(number2 % number1, number1);
  }
}
void YES(bool condition) {
  if (condition) {
    cout << "YES"
         << "\n";
    ;
  } else {
    cout << "NO"
         << "\n";
    ;
  }
  return;
}
void Yes(bool condition) {
  if (condition) {
    cout << "Yes"
         << "\n";
    ;
  } else {
    cout << "No"
         << "\n";
    ;
  }
  return;
}
long long n, m, num, sum, a, b, c, d, e, g, h, w, i, j, q, r, l;
long long k;
long long ans;
long long x[100005], y[500005], z[200005];
char s[500005];
long long dp[104][104][104];
long long query;
long long len[105];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> k;
  cin >> s;
  m = n;
  n = k;
  for (long long i = 0; i < m; i++) {
    if (s[i] == '1') {
      dp[1][i][i] = 1;
    } else {
      dp[0][i][i] = 1;
    }
  }
  len[0] = 2;
  len[1] = 2;
  for (long long i = 2; i <= n; i++) {
    len[i] = len[i - 1] * len[i - 2];
    len[i] %= mod;
  }
  for (long long i = 2; i <= n; i++) {
    for (long long j = 0; j < m; j++) {
      for (long long k = j; k < m; k++) {
        if (j == 0 && k == m - 1) {
          dp[i][j][k] +=
              dp[i - 1][j][k] * len[i - 2] + dp[i - 2][j][k] * len[i - 1];
        } else if (j == 0) {
          dp[i][j][k] += dp[i - 1][j][k] + dp[i - 2][j][k] * len[i - 1];
        } else if (k == m - 1) {
          dp[i][j][k] += dp[i - 2][j][k] + dp[i - 1][j][k] * len[i - 2];
        } else {
          dp[i][j][k] += dp[i - 1][j][k] + dp[i - 2][j][k];
        }
        dp[i][j][k] %= mod;
        for (long long l = j; l < k; l++) {
          dp[i][j][k] += dp[i - 1][j][l] * dp[i - 2][l + 1][k];
          dp[i][j][k] %= mod;
        }
        dp[i][j][k] %= mod;
      }
    }
  }
  ans = dp[n][0][m - 1];
  cout << ans << "\n";
  ;
  return 0;
}