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 long mod = 1e9 + 7;
const int MAXN = 102;
namespace matrix_ops {
template <typename T>
struct matrix_mod {
  vector<vector<T> > data;
  int n, m;
  matrix_mod(int _n, int _m, T v = T())
      : data(vector<vector<T> >(_n, vector<T>(_m, v))), n(_n), m(_m) {}
  matrix_mod operator*(matrix_mod &other) {
    matrix_mod res(n, other.m);
    for (int i = 0; i < n; i++) {
      for (int k = 0; k < m; k++) {
        for (int j = 0; j < other.m; j++) {
          res.data[i][j] += (data[i][k] * other.data[k][j]) % mod;
          if (res.data[i][j] >= mod) res.data[i][j] -= mod;
        }
      }
    }
    return res;
  }
};
template <typename T>
matrix_mod<T> get_diag_mat(int n, T val) {
  matrix_mod<T> res(n, n);
  for (int i = 0; i < n; i++) res.data[i] = val;
  return res;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) os << ", ";
    os << v[i];
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, matrix_mod<T> &mat) {
  os << "{\n";
  for (auto &row : mat.data) {
    os << '\t' << row << endl;
  }
  return os << "}";
}
}  // namespace matrix_ops
using namespace matrix_ops;
vector<int> prefix(const string &s) {
  vector<int> P(s.size(), 0);
  int k = 0;
  for (int i = 1; i < s.size(); i++) {
    while (k > 0 and s[i] != s[k]) k = P[k - 1];
    if (s[i] == s[k]) k++;
    P[i] = k;
  }
  return P;
}
vector<vector<int> > TT(const string &s, const vector<int> &P) {
  int sz = P.size();
  vector<vector<int> > res(sz, vector<int>(2, 0));
  auto move_forward = [&](int curr, char c) -> int {
    while (curr > 0 and s[curr] != c) curr = P[curr - 1];
    if (s[curr] == c) curr++;
    return curr == sz ? P[curr - 1] : curr;
  };
  for (int i = 0; i < sz; i++) {
    res[i][0] = move_forward(i, '0');
    res[i][1] = move_forward(i, '1');
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  using field = matrix_mod<long long>;
  int n, x;
  string s;
  cin >> n >> x;
  cin >> s;
  vector<int> P = prefix(s);
  vector<vector<int> > Fail = TT(s, P);
  vector<field> mats(MAXN, field(s.size() + 1, s.size() + 1));
  int last = s.back() - '0';
  for (int dig = 0; dig < 2; dig++) {
    for (int k = 0; k < s.size(); k++) {
      mats[dig].data[k][k]++;
      mats[dig].data[Fail[k][dig]][k]++;
    }
    mats[dig].data.back().back() += 2;
  }
  mats[last].data.back()[s.size() - 1]++;
  for (int i = 2; i < mats.size(); i++) {
    mats[i] = mats[i - 2] * mats[i - 1];
  }
  long long ans = mats[x].data.back().front();
  cout << ans << '\n';
  return 0;
}