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 <cstdio>
#include <iostream>
using namespace std;
const int maxn=21;

int f[1<<maxn][maxn],n,K;
char sx[1<<maxn];

int main(void)
{
    scanf("%d%d",&n,&K);
    for(int i=0;i<=n;++i)
    {
        scanf("%s",sx);
        for(int j=0;j<(1<<i);++j)
            f[j|(1<<i)][i]=sx[j]=='1';
    }
    for(int i=n;i>=0;--i)
        for(int s=0;s<(1<<i);++s)
            for(int j=i;j>=0;--j)
            {
                int tmp=f[(1<<i)|s][j];
                int s1=s>>j;
                int s2=s&((1<<j)-1);
                if(j) f[(1<<(i-j))|s1][0]+=tmp;
                if(s2)
                {
                    int p=j-1;while(!((s2>>p)&1)) --p;
                    f[(1<<(i-j+p+1))|(s1<<(p+1))|(1<<p)|(s2&((1<<p)-1))][p]+=tmp;
                }
                if(((1<<j)-1)&(~s2))
                {
                    int p=j-1;while((s2>>p)&1) --p;
                    f[(1<<(i-j+p+1))|(s1<<(p+1))|(s2&(1<<p)-1)][p]+=tmp;
                }
            }
    for(int i=n;i>=0;--i)
        for(int j=0;j<(1<<i);++j)
            if(f[(1<<i)|j][0]>=K)
            {
                for(int k=i-1;k>=0;--k) printf("%d",(j>>k)&1);
                return 0;
            }
    return 0;
}