//BadWaper gg
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<deque>
#include<bitset>
#include<map>
#include<set>
#define inf 1e9
#define eps 1e-6
#define mp make_pair
#define N 21
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline ll read()
{
	char ch=getchar();
	ll s=0,w=1;
	while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}
	return s*w;
}
int f[1<<N][N];
int n,m;
char s[1<<N];
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	n=read(),m=read();
	for(register int i=0;i<=n;i++)
	{
		scanf("%s",s);
		for(register int j=0;j<(1<<i);j++)
		{
			f[(1<<i)|j][i]=(s[j]-'0');
		}
	}
	for(register int i=n;i>=0;i--)
	{
		for(register int j=0;j<(1<<i);j++)
		{
			for(register int k=i;k;k--)
			{
				if(!f[(1<<i)|j][k])continue;
				int v=f[(1<<i)|j][k],S=(j>>k),T=(j&((1<<k)-1));
				f[(1<<(i-k))|S][0]+=v;
				int u1=(k-1),u2=(k-1);
				while((u1>=0)&&((T>>u1)&1))u1--;
				while((u2>=0)&&(!((T>>u2)&1)))u2--;
				if(u2>=0){f[(T&((1<<u2)-1))|(S<<(u2+1))|(1<<(i-k+u2+1))|(1<<u2)][u2]+=v;}
				if(u1>=0){f[(T&((1<<u1)-1))|(S<<(u1+1))|(1<<(i-k+u1+1))][u1]+=v;}
			}
		}
	}
	for(register int i=n;i>=0;i--)
	{
		for(register int j=0;j<(1<<i);j++)
		{
			if(f[(1<<i)|j][0]>=m)
			{
				for(register int k=i-1;k>=0;k--)printf("%d",(j>>k)&1);printf("\n");
				return 0;
			}
		}
	}
	puts("");
	return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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
//BadWaper gg
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<deque>
#include<bitset>
#include<map>
#include<set>
#define inf 1e9
#define eps 1e-6
#define mp make_pair
#define N 21
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline ll read()
{
	char ch=getchar();
	ll s=0,w=1;
	while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}
	return s*w;
}
int f[1<<N][N];
int n,m;
char s[1<<N];
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	n=read(),m=read();
	for(register int i=0;i<=n;i++)
	{
		scanf("%s",s);
		for(register int j=0;j<(1<<i);j++)
		{
			f[(1<<i)|j][i]=(s[j]-'0');
		}
	}
	for(register int i=n;i>=0;i--)
	{
		for(register int j=0;j<(1<<i);j++)
		{
			for(register int k=i;k;k--)
			{
				if(!f[(1<<i)|j][k])continue;
				int v=f[(1<<i)|j][k],S=(j>>k),T=(j&((1<<k)-1));
				f[(1<<(i-k))|S][0]+=v;
				int u1=(k-1),u2=(k-1);
				while((u1>=0)&&((T>>u1)&1))u1--;
				while((u2>=0)&&(!((T>>u2)&1)))u2--;
				if(u2>=0){f[(T&((1<<u2)-1))|(S<<(u2+1))|(1<<(i-k+u2+1))|(1<<u2)][u2]+=v;}
				if(u1>=0){f[(T&((1<<u1)-1))|(S<<(u1+1))|(1<<(i-k+u1+1))][u1]+=v;}
			}
		}
	}
	for(register int i=n;i>=0;i--)
	{
		for(register int j=0;j<(1<<i);j++)
		{
			if(f[(1<<i)|j][0]>=m)
			{
				for(register int k=i-1;k>=0;k--)printf("%d",(j>>k)&1);printf("\n");
				return 0;
			}
		}
	}
	puts("");
	return 0;
}

```