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<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
int read()
{
	char ch=getchar();int f=0,x=1;
	while(ch<'0'||ch>'9'){if(ch=='-') x=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){f=(f<<1)+(f<<3)+ch-'0';ch=getchar();}
	return f*x;
}
int a[2200005];
int f[23][2200005],n,k,g[23][2200005],m,len,ans;
char s[2000005];
int main()
{
	n=read();m=read();
	for(int i=0;i<=n;i++)
	{
		scanf("%s",s);
		int x=strlen(s);
		for(int j=0;j<x;j++)
		{
			if(s[j]=='1')
			f[i][j]=1;
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<(1<<i);j++)
		{
			int b=j>>(i-1)&1;int k;
			for(k=0;k<i&&(j>>i-k-1&1)==b;k++);
			g[i][j]=i==k?-1:k;
		}
	}
	for(int i=0;i<=n;i++)
	{
		for(int j=1;i+j<=n;j++)
		{
			for(int k=0;k<1<<i+j;k++)
			{
				f[i][k>>j]+=f[i+j][k];
				int t=g[j][k&(1<<j)-1];
				if (~t)
				{
					f[i+j-t][(k>>j<<j-t)|(k&(1<<j-t)-1)]+=f[i+j][k];
				}
			}
		}
		for(int j=(1<<i)-1;~j;--j)
		{
    		if(f[i][j]>=m)
			{
				len=i,ans=j;
    		}
		}
	}
	for(int i=len-1;~i;--i)
	{
		putchar((ans>>i&1)+'0');
	}
}