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<bits/stdc++.h>
using namespace std;
const int N=22;
int ch[2][1<<N],id[N][1<<(N-1)],cnt,a[N][1<<N],siz[N],c[N][1<<N],num[N][1<<N],ans[N],n,K;
char s[1<<(N-1)];
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch))
	{
		if(ch=='-') f=-1;
		ch=getchar();
	}
	while(isdigit(ch))
	{
		x=(x<<1)+(x<<3)+ch-'0';
		ch=getchar();
	}
	return x*f;
}
inline void dfs(int d,int s)
{
	if(ans[d]==-1)ans[d]=s;
	for(int p=0;p<=1;p++)
	{
		int tot=0;
		for(int i=0;i<siz[d];i++)
		{
			int v=ch[p][a[d][i]];
			if(v==-1) continue;
			if(num[d+1][v]==-1)
			{
				a[d+1][siz[d+1]]=v;
				num[d+1][v]=siz[d+1];
				siz[d+1]++;
			}
			tot+=c[d][i];
			c[d+1][num[d+1][v]]+=c[d][i];
		}
		if(tot>=K)dfs(d+1,(s<<1)|p);
		for(int i=0;i<siz[d+1];i++)
		{
			num[d+1][a[d+1][i]]=-1;
			c[d+1][i]=0;
		}
		siz[d+1]=0;
	}
}
inline void pre()
{
	id[0][0]=(cnt++);
	ch[0][id[0][0]]=ch[1][id[0][0]]=-1;
	for(int i=1;i<=20;i++)
	{
		int t=(1<<i),k=(t>>1);
		for(int s=0;s<t;s++)
		{
			id[i][s]=(cnt++);
			int p=(s&k),q=(p?1:0);
			ch[q][id[i][s]]=id[i-1][s^p];
			ch[!q][id[i][s]]=ch[!q][id[i-1][s^p]];
		}
	}
}
int main(void)
{
	n=read();
	K=read();
	pre();
	for(int i=0;i<=n;i++)for(int j=0;j<cnt;j++)num[i][j]=-1;
	for(int i=0;i<=n;i++)
	{
		scanf("%s",s);
		int t=(1<<i);
		for(int j=0;j<t;j++)if(s[j]=='1')
		{
			num[0][a[0][siz[0]]=id[i][j]]=siz[0];
			c[0][num[0][id[i][j]]]++;
			siz[0]++;
		}
		ans[i]=-1;
	}
	dfs(0,0);
	for(int i=n;i>=0;i--)if(ans[i]!=-1)
	{
		for(int j=i-1;j>=0;j--)putchar(((ans[i]>>j)&1)+'0');
		puts("");
		return 0;
	}
}