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
#include <bits/stdc++.h>
using namespace std;
long long mod = (1000000007LL);
inline long long Mod(long long a, long long b) { return (a % b); }
inline long long poww(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b, swap(a, b);
  }
  return a;
}
void read(vector<long long> &w, long long n) {
  w.resize(n);
  for (long long i = 0; i < n; i++) cin >> w[i];
}
void print(vector<long long> &w) {
  for (long long i = 0; i < (long long)(w).size(); i++) {
    if (i == (long long)(w).size() - 1)
      cout << w[i] << "\n";
    else
      cout << w[i] << " ";
  }
}
long long prodmod(vector<long long> w);
long long summod(vector<long long> w);
long long n, m, q, k, v[110], ans, cost[110];
pair<long long, long long> w[110];
string second;
long long dp[110][110][110];
long long solve(long long ini, long long fim, long long bloco) {
  if (ini > fim) return 0;
  if (bloco < 2) {
    if (second[ini] - '0' == bloco and ini == fim) return 1;
    return 0;
  }
  if (dp[ini][fim][bloco] != -1) return dp[ini][fim][bloco];
  long long ans = 0;
  long long custo1 = (ini == 0 ? cost[bloco - 1] : 1),
            custo2 = (fim == n - 1 ? cost[bloco - 2] : 1);
  ans += (solve(ini, fim, bloco - 2) * custo1) % mod;
  ans += (solve(ini, fim, bloco - 1) * custo2) % mod;
  ans %= mod;
  for (long long k = ini; k < fim; k++) {
    ans +=
        (1LL * solve(ini, k, bloco - 1) * solve(k + 1, fim, bloco - 2)) % mod;
    ans %= mod;
  }
  return dp[ini][fim][bloco] = ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  cin >> second;
  cost[0] = cost[1] = 2LL;
  for (long long i = 2; i < 110; i++)
    cost[i] = (cost[i - 2] * cost[i - 1]) % mod;
  memset(dp, -1, sizeof dp);
  cout << solve(0, n - 1, k) << "\n";
}
long long summod(vector<long long> w) {
  long long curr = 0;
  for (long long i = 0; i < (long long)(w).size(); i++) {
    curr = (curr + w[i]) % mod;
    if (curr < 0) curr += mod;
  }
  return curr;
}
long long prodmod(vector<long long> w) {
  long long curr = 1;
  for (long long i = 0; i < (long long)(w).size(); i++) {
    if (w[i] >= mod) w[i] %= mod;
    curr = (curr * w[i]) % mod;
    if (curr < 0) curr += mod;
  }
  return curr;
}
