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>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int f[21][1<<21];
int to[1<<21][2];
char S[1<<21];
int val[21][1<<20];
int main()
{
	int n,k;
	scanf("%d %d",&n,&k);
	to[1][0]=to[1][1]=0;
	for(int i=2;i<(1<<(n+1));i++)
	{
		to[i][i&1]=i>>1;
		to[i][(i&1)^1]=to[i>>1][(i&1)^1];
	}
	for(int i=0;i<=n;i++)
	{
		scanf("%s",S);
		for(int x=0;x<(1<<i);x++)
		{
			if(S[x]=='0')continue;
			int y=0;
			for(int j=0;j<i;j++)
				if(x&(1<<j))
					y|=1<<(i-j-1);
			y|=(1<<i);
			f[0][y]++;
		}
	}
	for(int i=0;i<n;i++)
	{
		for(int s=0;s<(1<<i);s++)
			for(int t=1;t<(1<<(n-i+1));t++)
			{
				f[i+1][s|(to[t][0]<<(i+1))]+=f[i][s|(t<<i)];
				f[i+1][(s|(1<<i))|(to[t][1]<<(i+1))]+=f[i][s|(t<<i)];
			}
	}
	for(int i=0;i<=n;i++)
		for(int s=0;s<(1<<i);s++)
			for(int t=1;t<(1<<(n-i+1));t++)
				val[i][s]+=f[i][s|(t<<i)];
	for(int i=n;i>=0;i--)
	{
		int res=1e9;
		for(int s=0;s<(1<<i);s++)
		{
			if(val[i][s]<k)continue;
			int x=0;
			for(int j=0;j<i;j++)
				if(s&(1<<j))
					x|=1<<(i-j-1);
			res=min(res,x);
		}
		if(res>(1<<i))continue;
		for(int j=i-1;j>=0;j--)
			printf("%d",(res>>j)&1);
		printf("\n");
		return 0;
	}
}