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<cstring>

namespace fk
{
	const int N=21;

	int t0[1<<N][N];

	int suf[N+1];

	void initialize()
	{
		for(int i=N-1;i;i--)
			suf[i]=suf[i+1]|(1<<i);
	}
	
	inline int& f(int S,int len,int ind)
	{
		return t0[S|suf[len+1]][ind];
	}

	int err_val=-1;
	int& to(int S,int len,int ind,bool rag)
	{
		if(ind==len)return err_val;

		if(((S>>ind)&1)==rag)
		{
			return f(S,len,ind+1);
		}
		else 
		{
			int fk=(1<<ind)-1,a=S&fk,b=S>>(ind+1);

//			bool flag=0;
//			if(S==2 && len==4 && ind==0)flag=1;

//			if(flag)printf("fuck : %d %d %d\n",fk,a,b);

			S=a|(b<<ind);

//			if(flag)printf("S --> %d\n",S);

			return to(S,len-1,ind,rag);
		}
	}

	char str[1<<N];

	int n,m;

	void output(int S,int len)
	{
		for(int i=0;i<len;i++)
			printf("%d",(S>>i)&1);
	}

	void solve()
	{
		initialize();

		scanf("%d%d",&n,&m);

		int ept;
		scanf("%d",&ept);
		for(int i=1;i<=n;i++)
		{
			scanf("%s",str);
			for(int j=0;j<(1<<i);j++)
				if(str[j]=='1')
				{
					int tmp=0;
					for(int k=0;k<i;k++)
						tmp|=((j>>k)&1)<<(i-k-1);

//					output(tmp,i),printf("\n");

//					printf("at begining : f["),output(tmp,i);
//					printf("][%d] = %d\n",0,1);

					f(tmp,i,0)=1;
				}
		}

		for(int len=n;len;len--)
			for(int i=0;i<(1<<len);i++)
				for(int ind=0;ind<=len;ind++)
				{
					int v=f(i,len,ind);

//					if(len==4 && i==2)
//					{
//						printf("f["),output(i,len),printf("][%d] = %d\n",ind,v);
//					}

					for(int rag=0;rag<2;rag++)
					{
						int &w=to(i,len,ind,rag);


//						if(len==4 && i==2)
//						{
//							printf("%d:::%d\n",rag,w);
//						}

						if(w==-1)continue;

						w+=v;
					}

					if(ind>0 && ind<len)
					{
						int &w=f(i&((1<<ind)-1),ind,ind);

						w+=v;
					}
				}

//		printf("%d\n",f(1,3,0));
//		printf("%d\n",f(1,3,1));

		int ans=-1,anslen=0; 

		for(int len=n;len;len--)
		{
			for(int i=0;i<(1<<len);i++)
			{

				int j=0;
				for(int k=0;k<len;k++)
					j|=((i>>k)&1)<<(len-k-1);

//				output(i,len);
//				printf(" ---> %d\n",f(i,len,len));

				if(f(j,len,len)>=m)
				{
					ans=j,anslen=len;
					goto out;
				}
			}
		}

//		printf("anslen = %d\n",anslen);

out:	output(ans,anslen);
		printf("\n");
	}

}

int main()
{
//	freopen("F.in","r",stdin);

	fk::solve();
	return 0;
}