#include<bits/stdc++.h>
#define MN 21
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
char st[1050005];
int n,f[1<<MN][MN+1],K,nxt0[1<<MN],nxt1[1<<MN],p[1<<MN],Len,from,sum,g[1<<MN][2];
void dfs(int x,int s,int v)
{
	if(v<K) return;
	if(x>Len) Len=x,from=s;
	else if(x==Len) from=min(from,s);
	dfs(x+1,s<<1,v-g[s|(1<<x-1)][0]);
	dfs(x+1,s<<1|1,v-g[s|(1<<x-1)][1]);
}
int main()
{
	n=read();K=read();
	for(int i=0;i<=n;++i)
	{
		scanf("%s",st);
		for(int j=0;j<1<<i;++j) 
		{
			p[j|(1<<i)]=j;
			if(st[j]=='1') ++f[j|(1<<i)][i],++sum;
		}
	}
	int tot=1<<n+1;
	for(int j=1,hbit=0;j<tot;++j)
	{
		if(j>=(1<<hbit+1)) ++hbit;
		int nx0=-1,nx1=-1;
		for(int k=hbit-1;k>=0;--k)
		{
			if(nx0<0&&!(j&(1<<k))) nx0=k;	
			if(nx1<0&& (j&(1<<k))) nx1=k;
		}
		nxt0[j]=nx0;nxt1[j]=nx1;
	}
	for(int i=tot;--i;) for(int j=n;~j;--j) if(f[i][j])
	{
		int r=(i&((1<<j)-1))|(1<<j);
		if(nxt0[r]>=0) f[((i>>j)<<nxt0[r]+1)|(i&((1<<nxt0[r]+1)-1))][nxt0[r]]+=f[i][j];
		else g[i>>j][0]+=f[i][j];
		if(nxt1[r]>=0) f[((i>>j)<<nxt1[r]+1)|(i&((1<<nxt1[r]+1)-1))][nxt1[r]]+=f[i][j];
		else g[i>>j][1]+=f[i][j];
	}
	dfs(1,0,sum);
	for(int i=Len-2;i>=0;--i) printf("%d",bool(from&(1<<i)));
	return 0;
}


### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
#define MN 21
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
char st[1050005];
int n,f[1<<MN][MN+1],K,nxt0[1<<MN],nxt1[1<<MN],p[1<<MN],Len,from,sum,g[1<<MN][2];
void dfs(int x,int s,int v)
{
	if(v<K) return;
	if(x>Len) Len=x,from=s;
	else if(x==Len) from=min(from,s);
	dfs(x+1,s<<1,v-g[s|(1<<x-1)][0]);
	dfs(x+1,s<<1|1,v-g[s|(1<<x-1)][1]);
}
int main()
{
	n=read();K=read();
	for(int i=0;i<=n;++i)
	{
		scanf("%s",st);
		for(int j=0;j<1<<i;++j) 
		{
			p[j|(1<<i)]=j;
			if(st[j]=='1') ++f[j|(1<<i)][i],++sum;
		}
	}
	int tot=1<<n+1;
	for(int j=1,hbit=0;j<tot;++j)
	{
		if(j>=(1<<hbit+1)) ++hbit;
		int nx0=-1,nx1=-1;
		for(int k=hbit-1;k>=0;--k)
		{
			if(nx0<0&&!(j&(1<<k))) nx0=k;	
			if(nx1<0&& (j&(1<<k))) nx1=k;
		}
		nxt0[j]=nx0;nxt1[j]=nx1;
	}
	for(int i=tot;--i;) for(int j=n;~j;--j) if(f[i][j])
	{
		int r=(i&((1<<j)-1))|(1<<j);
		if(nxt0[r]>=0) f[((i>>j)<<nxt0[r]+1)|(i&((1<<nxt0[r]+1)-1))][nxt0[r]]+=f[i][j];
		else g[i>>j][0]+=f[i][j];
		if(nxt1[r]>=0) f[((i>>j)<<nxt1[r]+1)|(i&((1<<nxt1[r]+1)-1))][nxt1[r]]+=f[i][j];
		else g[i>>j][1]+=f[i][j];
	}
	dfs(1,0,sum);
	for(int i=Len-2;i>=0;--i) printf("%d",bool(from&(1<<i)));
	return 0;
}


```