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 N = 1e9 + 7;
long long power(long long a, long long b, long long mod) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2) {
      x = (x * y) % mod;
    }
    y = (y * y) % mod;
    b /= 2;
  }
  return x % mod;
}
map<string, long long> M;
long long fib[105];
long long stn[20000];
void mpas(string s) {
  long long a, b, c;
  string y;
  long long o;
  a = s.length();
  long long j = 1;
  long long i, k;
  k = 1;
  for (i = 0; i < a; i++) {
    for (j = i; j < a; j++) {
      y = s.substr(i, j - i + 1);
      o = i + 101 * j;
      if (M.count(y)) {
        stn[o] = M[y];
        continue;
      }
      M[y] = k++;
      stn[o] = M[y];
    }
  }
}
long long dp[105][10005];
long long st[105][10005], en[105][10005];
long long ns[105][10005];
long long nos(long long a, long long o) {
  long long b, c, d, i, j, k, n, x;
  long long e, f;
  string p, q, r;
  b = stn[o];
  if (ns[a][b] != -1) return ns[a][b];
  d = 0;
  d += (nos(a - 1, o) + nos(a - 2, o)) % N;
  long long o1 = o % 101;
  long long o2 = o / 101;
  c = o2 - o1 + 1;
  for (i = 1; i <= c - 1; i++) {
    e = o1 + 101 * (o1 + i - 1);
    f = (o1 + i) + 101 * (o2);
    d += (nos(a - 1, e) * nos(a - 2, f)) % N;
    d %= N;
  }
  return ns[a][b] = d;
}
long long sta(long long a, long long o) {
  long long b, c, d, i, j, k, n, x, e, f;
  string p, q, r;
  b = stn[o];
  if (st[a][b] != -1) return st[a][b];
  long long o1 = o % 101;
  long long o2 = o / 101;
  c = o2 - o1 + 1;
  d = 0;
  d += (sta(a - 1, o) * (power(2, fib[a - 2], N))) % N;
  d %= N;
  d += sta(a - 2, o);
  d %= N;
  for (i = 1; i <= c - 1; i++) {
    e = o1 + 101 * (o1 + i - 1);
    f = (o1 + i) + 101 * (o2);
    d += (nos(a - 1, e) * sta(a - 2, f)) % N;
    d %= N;
  }
  return st[a][b] = d;
}
long long end1(long long a, long long o) {
  long long b, c, d, i, j, k, n, x, e, f;
  string p, q, r;
  b = stn[o];
  if (en[a][b] != -1) return en[a][b];
  long long o1 = o % 101;
  long long o2 = o / 101;
  c = o2 - o1 + 1;
  d = 0;
  d += (end1(a - 2, o) * (power(2, fib[a - 1], N))) % N;
  d %= N;
  d += end1(a - 1, o);
  d %= N;
  for (i = 1; i <= c - 1; i++) {
    e = o1 + 101 * (o1 + i - 1);
    f = (o1 + i) + 101 * (o2);
    d += (end1(a - 1, e) * nos(a - 2, f)) % N;
    d %= N;
  }
  return en[a][b] = d;
}
long long fn(long long a, long long o) {
  long long b, c, d, i, j, k, n, x;
  long long e, f;
  string p, q, r;
  b = stn[o];
  if (dp[a][b] != -1) return dp[a][b];
  long long o1 = o % 101;
  long long o2 = o / 101;
  c = o2 - o1 + 1;
  long long ans = 0;
  ans += ((fn(a - 1, o) * (power(2, fib[a - 2], N))) % N +
          (fn(a - 2, o) * (power(2, fib[a - 1], N))) % N) %
         N;
  ans %= N;
  for (i = 1; i <= c - 1; i++) {
    e = o1 + 101 * (o1 + i - 1);
    f = (o1 + i) + 101 * (o2);
    ans += (end1(a - 1, e) * sta(a - 2, f)) % N;
    ans %= N;
  }
  return dp[a][b] = ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(25);
  long long a, b, c, d, i, j, k, n, x;
  string p, q, r, s;
  fib[0] = fib[1] = 1;
  for (i = 2; i <= 100; i++) fib[i] = (fib[i - 1] + fib[i - 2]) % (N - 1);
  cin >> n >> x >> s;
  mpas(s);
  for (i = 0; i < 105; i++)
    for (j = 0; j < 10005; j++)
      dp[i][j] = -1, st[i][j] = -1, en[i][j] = -1, ns[i][j] = -1;
  for (i = 0; i < 10005; i++)
    dp[0][i] = dp[1][i] = st[0][i] = en[1][i] = st[1][i] = en[0][i] = 0,
    ns[0][i] = ns[1][i] = 0;
  if (M.count("0"))
    dp[0][M["0"]] = 1, st[0][M["0"]] = 1, en[0][M["0"]] = 1, ns[0][M["0"]] = 1;
  if (M.count("1"))
    dp[1][M["1"]] = 1, st[1][M["1"]] = 1, en[1][M["1"]] = 1, ns[1][M["1"]] = 1;
  cout << fn(x, 101 * (n - 1)) << "\n";
  return 0;
}
