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:
//传说中的序列自动机 get √ 
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
int n,K,dp[1<<21][21],nxt[1<<21][2],Log[1<<21];
bool inq[1<<21][21];
struct node
{
	int S,pos;
};
queue<node>q;
void init()
{
	for(int S=1;S<(1<<n);S++)
	{
		Log[S]=log2(S);
		nxt[S][0]=nxt[S][1]=-1;
		for(int i=0,f0=0,f1=0;i<Log[S]&&(!f0||!f1);i++)
			if(S&(1<<i))
			{
				if(!f1)nxt[S][1]=S>>i;
				f1=1;
			}
			else
			{
				if(!f0)nxt[S][0]=S>>i;
				f0=1;
			}
	}
}
void bfs()
{
	while(!q.empty())
	{
		int S=q.front().S,pos=q.front().pos;
		q.pop();
		if(pos==Log[S])continue;
		int L=S&((1<<pos)-1),R=S>>pos;
		int T=L+(1<<pos),pos2=pos;
		dp[T][pos2]+=dp[S][pos];
		if(!inq[T][pos2])
			inq[T][pos2]=1,q.push((node){T,pos2});
		if(nxt[R][0]!=-1)
		{
			T=L+(nxt[R][0]<<pos);
			pos2=pos+1;
			dp[T][pos2]+=dp[S][pos];
			if(!inq[T][pos2])
				inq[T][pos2]=1,q.push((node){T,pos2});
		}
		if(nxt[R][1]!=-1)
		{
			T=L+(nxt[R][1]<<pos);
			pos2=pos+1;
			dp[T][pos2]+=dp[S][pos];
			if(!inq[T][pos2])
				inq[T][pos2]=1,q.push((node){T,pos2});
		}
	}
	int maxn=-1,ans=0;
	for(int S=1;S<(1<<n);S++)
		if(dp[S][Log[S]]>=K&&Log[S]>maxn)
			maxn=Log[S],ans=S;
	for(int i=Log[ans]-1;i>=0;i--)
		printf("%d",(ans>>i)&1);
	puts("");
}
int main()
{
	scanf("%d %d",&n,&K),n++;
	init();
	for(int i=1;i<=n;i++)
		for(int j=1,x;j<=(1<<(i-1));j++)
		{
			scanf("%1d",&x);
			if(x==1)
			{
				int S=j-1+(1<<(i-1));
				dp[S][0]++;
				if(!inq[S][0])
					inq[S][0]=1,q.push((node){S,0});
			}
		}
	bfs();
	return 0;
}
/*
3 4
1
01
1011
01001110

10
*/