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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class F946{

    long[] fibpow;
    char[] chs;
    long mod = 1000000007L;
    long modpow = mod-1L;
    long[][][] memo;

    long pow(long m, long n){
        long res = 1;
        while(n > 0){
            if(n % 2 == 1)res = (res*m)%mod;
            m = (m*m)%mod;
            n = n/2;
        }
        return res;
    }


    private void solve() throws Exception {
        int n = nextInt();
        int x = nextInt();
        chs = nextToken().toCharArray();
        memo = new long[n][n][x+1];
        for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) Arrays.fill(memo[i][j], -1);
        fibpow = new long[x+1];
        if(x > 0) fibpow[1] = 1L;
        for(int i = 2; i <= x; i++) fibpow[i] = (fibpow[i-1]+ fibpow[i-2])%modpow;
        for(int i = 0; i <= x; i++) fibpow[i] = pow(2, fibpow[i]);
        out.println(doit(0,n-1,x));
    }

    long doit(int i, int j, int x) {
        if (x < 2 && i == j && chs[i]-'0' == x) return 1L;
        else if (x < 2 && i == j && chs[i]-'0' != x) return 0L;
        else if(x < 2 && i < j) return 0L;
        else if(memo[i][j][x] < 0){
            long res = 0L;
            res = (res + ((i==0?fibpow[x]:(1L)) * doit(i,j, x-2))%mod)%mod;
            for(int t = i; t < j; t++) res = (res + (doit(i,t,x-1) * doit(t+1,j, x-2))%mod)%mod;
            res = (res + (doit(i,j,x-1) * (j==chs.length-1? fibpow[x-1]:(1L)))%mod)%mod;
            memo[i][j][x] = res;
        }
        return memo[i][j][x];
    }


    public static void main(String[] args) {
        (new F946()).run();
    }

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer tokenizer;

    public void run() {
        try {
            in = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            out = new PrintWriter(System.out);
            solve();
            in.close();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private float nextFloat() throws IOException {
        return Float.parseFloat(nextToken());
    }

    private String nextLine() throws IOException {
        return new String(in.readLine());
    }

    private String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(in.readLine());
        }
        return tokenizer.nextToken();
    }


}