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:
#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
#include <iostream>
#include <algorithm>
#include <cstdio>
#define N 21
using namespace std;

int a[N][1 << N | 1];
int f[1 << N | 1][2];

inline int L(int x)
{
    return 31 - __builtin_clz(x);
}

int main(void)
{
    int n, k;
    int i, j, p, q;

    scanf("%d %d", &n, &k);
    for(i = 0; i <= n; i ++)
        for(j = 0; j < (1 << i); j ++)
            scanf("%1d", &a[i][j | 1 << i]);

    for(i = 2; i < (1 << (n + 1)); i ++)
        if(i & (1 << (L(i) - 1)))
        {
            f[i][1] = i ^ (1 << L(i));
            f[i][0] = f[f[i][1]][0];
        }
        else
        {
            f[i][0] = i ^ (1 << L(i)) ^ (1 << (L(i) - 1));
            f[i][1] = f[f[i][0]][1];
        }

    for(i = n; i >= 1; i --)
        for(j = 1 << i; j < (1 << (n + 1)); j ++)
            if(a[i][j])
            {
                p = j >> i;
                q = (j & ((1 << i) - 1)) | (1 << i);
                if(f[q][0])
                    a[L(f[q][0])][(p << 1 | 0) << L(f[q][0]) | (f[q][0] ^ (1 << L(f[q][0])))] += a[i][j];
                if(f[q][1])
                    a[L(f[q][1])][(p << 1 | 1) << L(f[q][1]) | (f[q][1] ^ (1 << L(f[q][1])))] += a[i][j];
                a[0][p] += a[i][j];
            }
    // for(i=1;i<(1<<(n+1));i++)cout<<a[0][i]<<' ';cout<<endl;

    for(i = p = 1; i < (1 << (n + 1)); i ++)
        if(a[0][i] >= k && L(i) > L(p))
            p = i;
    for(i = L(p) - 1; i > -1; i --)
        printf("%d", !!(p & (1 << i)));
    printf("\n");

    return 0;
}