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>
using namespace std;
#define N 22
int dp[1<<N][N],n,k,v[N],nt[N][2],ct,le,lg[1<<N];
char s[1<<N];
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=0;i<=n;i++)
	{
		scanf("%s",s);
		for(int j=0;j<1<<i;j++)
		dp[j|(1<<i)][0]=s[j]=='1';
	}
	for(int i=2;i<1<<n+1;i++)lg[i]=lg[i>>1]+1;
	for(int j=(1<<n+1)-1;j>=0;j--)
	{
		int le=lg[j];
		for(int k=1;k<=le;k++)v[k]=(bool)(j&(1<<k-1));
		nt[le][0]=nt[le][1]=-1;
		for(int k=le-1;k>=0;k--)
		{
			nt[k][0]=nt[k+1][0];nt[k][1]=nt[k+1][1];
			nt[k][v[k+1]]=k+1;
		}
		for(int k=0;k<=le;k++)
		if(dp[j][k])
		{
			dp[j&((1<<k)-1)][k+1]=dp[j&((1<<k)-1)][k+1]+dp[j][k];
			if(nt[k][0]!=-1)dp[(j&((1<<k)-1))|(j>>(nt[k][0]-1)<<k)][k+1]=dp[(j&((1<<k)-1))|(j>>(nt[k][0]-1)<<k)][k+1]+dp[j][k];
			if(nt[k][1]!=-1)dp[(j&((1<<k)-1))|(j>>(nt[k][1]-1)<<k)][k+1]=dp[(j&((1<<k)-1))|(j>>(nt[k][1]-1)<<k)][k+1]+dp[j][k];
		}
	}
	for(int l=n;l>=1;l--)
	for(int j=0;j<1<<l;j++)
	if(dp[j][l+1]>=k)
	{
		for(int i=l;i>=1;i--)printf("%d",(bool)(j&(1<<i-1)));
		return 0;
	}
}//orzzyw