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;
const long double eps = 1e-7;
const long double PI = acos(-1);
const int INF = (int)1e9;
const long long INFF = (long long)1e18;
const int mod = (int)1e9 + 7;
const int MXN = (int)1e2 + 7;
int dp[MXN][MXN][MXN];
int len[MXN];
char s[MXN];
int mypow(int v1, int t) {
  int res = 1;
  while (t) {
    if (t & 1) res = res * 1ll * v1 % mod;
    v1 = v1 * 1ll * v1 % mod;
    t >>= 1;
  }
  return res;
}
inline void add(int &v1, int v2) {
  v1 += v2;
  if (v1 >= mod) v1 -= mod;
  if (v1 < 0) v1 += mod;
}
int main(void) {
  int l, n;
  scanf("%d %d", &l, &n);
  scanf("%s", s + 1);
  for (int i = 1; i < l + 1; ++i) {
    if (s[i] == '1')
      dp[1][i][i] = 1;
    else
      dp[0][i][i] = 1;
  }
  len[0] = len[1] = 1;
  for (int i = 2; i < n + 1; ++i) {
    for (int ll = 1; ll < l + 1; ++ll)
      for (int rr = ll; rr < l + 1; ++rr) {
        int bs = 1;
        if (rr == l) bs = mypow(2, len[i - 2]);
        add(dp[i][ll][rr], dp[i - 1][ll][rr] * 1ll * bs % mod);
        bs = 1;
        if (ll == 1) bs = mypow(2, len[i - 1]);
        add(dp[i][ll][rr], dp[i - 2][ll][rr] * 1ll * bs % mod);
      }
    for (int ll = 1; ll < l + 1; ++ll)
      for (int md = ll; md < l + 1; ++md)
        for (int rr = md + 1; rr < l + 1; ++rr) {
          add(dp[i][ll][rr],
              dp[i - 1][ll][md] * 1ll * dp[i - 2][md + 1][rr] % mod);
        }
    len[i] = (len[i - 1] + len[i - 2]) % (mod - 1);
  }
  printf("%d\n", dp[n][1][l]);
  return 0;
}