Problem: You are given a set S of strings consisting of `0` and `1`, and an integer K.

Find the longest string that is a subsequence of K or more different strings in S. If there are multiple strings that satisfy this condition, find the lexicographically smallest such string.

Here, S is given in the format below:

* The data directly given to you is an integer N, and N+1 strings X_0,X_1,...,X_N. For every i (0\leq i\leq N), the length of X_i is 2^i.
* For every pair of two integers (i,j) (0\leq i\leq N,0\leq j\leq 2^i-1), the j-th character of X_i is `1` if and only if the binary representation of j with i digits (possibly with leading zeros) belongs to S. Here, the first and last characters in X_i are called the 0-th and (2^i-1)-th characters, respectively.
* S does not contain a string with length N+1 or more.



Here, a string A is a subsequence of another string B when there exists a sequence of integers t_1 < ... < t_{|A|} such that, for every i (1\leq i\leq |A|), the i-th character of A and the t_i-th character of B is equal.

Constraints

* 0 \leq N \leq 20
* X_i(0\leq i\leq N) is a string of length 2^i consisting of `0` and `1`.
* 1 \leq K \leq |S|
* K is an integer.

Input

Input is given from Standard Input in the following format:


N K
X_0
:
X_N


Output

Print the lexicographically smallest string among the longest strings that are subsequences of K or more different strings in S.

Examples

Input

3 4
1
01
1011
01001110


Output

10


Input

4 6
1
01
1011
10111010
1101110011111101


Output

100


Input

2 5
0
11
1111


Output

Solution (in CPP):

Solution:
#include<iostream>
using namespace std;
int n, k, i, j, ii, p, x, y;
int ad[22], d[(1 << 21) + 2][22];
char c;
int main(){
    cin>> n >> k;
    for(i = 1; i <= n; i++){
        ad[i] = ad[i - 1] + (1 << (i - 1) );
    }
    for(i = 0; i <= n; i++){
        for(ii = 0; ii < (1 << i); ii++){
            cin>> c;
            d[ii + ad[i] ][0] = c - '0';
        }
    }
    for(i = n; i >= 1; i--){
        for(j = 0; j < i; j++){
            for(ii = 0; ii < (1 << i); ii++){
                x = (ii >> (i - j) );
                d[x + ad[j] ][j] += d[ii + ad[i] ][j];
                for(p = j; p < i; p++){
                    if( ( (ii >> (i - p - 1) ) & 1) == 0){
                        y = x * 2 * (1 << (i - p - 1) ) + (ii & ( (1 << (i - p - 1) ) - 1) );
                        d[ y + ad[j + i - p] ][j + 1] += d[ii + ad[i] ][j];
                        break;
                    }
                }
                for(p = j; p < i; p++){
                    if( ( (ii >> (i - p - 1) ) & 1) == 1){
                        y = (x * 2 + 1) * (1 << (i - p - 1) ) + (ii & ( (1 << (i - p - 1) ) - 1) );
                        d[ y + ad[j + i - p] ][j + 1] += d[ii + ad[i] ][j];
                        break;
                    }
                }
            }
        }
        for(ii = 0; ii < (1 << i); ii++){
            if(d[ ii + ad[i] ][i] >= k){
                for(j = i - 1; j >= 0; j--){
                    cout<< ( (ii >> j) & 1);
                }
                return 0;
            }
        }
    }
}