#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,k,f[21][1<<20],ans,l;
char s[2000005];
void print(int x,int l)
{
	for(int i=l-1;i>=0;i--)
	  printf("%d",(x>>i)&1);
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=0;i<=n;i++)
	{
		scanf("%s",s);
		for(int j=0;j<(1<<i);j++)
		  if(s[j]=='1')f[i][j]=1;
	}
	for(int i=0;i<=n;i++)
	{
		for(int j=1;j<=n-i;j++)
		{
			for(int x=0;x<(1<<i);x++)
			for(int y=0;y<(1<<j);y++)
			{
				if(!f[j][(x<<j)+y])continue;
				f[0][x]+=f[j][(x<<j)+y];
//				print(x,i);
//				printf(" ");
//				print(y,j);
//				printf(" %d,val=%d\n",f[j][(x<<j)+y],f[0][0]);
			}
		}
		for(int x=(1<<i)-1;x>=0;x--)
		{
//				print(x,i);
//				printf(" %d\n",f[0][x]);
			if(f[0][x]>=k)ans=x,l=i; 
			f[0][x]=0;
		}
		for(int j=1;j<=n-i;j++)
		{
			for(int x=0;x<(1<<i);x++)
			for(int y=0;y<(1<<j);y++)
			{
				if(!f[j][(x<<j)+y])continue;
//				printf("f[");
//				print(x,i);
//				printf("][");
//				print(y,j);
//				printf("]=%d\n",f[j][(x<<j)+y]);
				int now=j-1,v=y;
				while(now>=0&&((1<<now)&y))v-=(1<<now),now--;
				if(now>=0)
				{
					f[now][(x<<(now+1))+v]+=f[j][(x<<j)+y];
//					print(x<<1,i+1);
//					printf(" ");
//					print(v,now);
//					printf("\n");
				}
				now=j-1,v=y;
				while(now>=0&&!((1<<now)&y))now--;
				if(now>=0)
				{
					f[now][(x<<(now+1))+v]+=f[j][(x<<j)+y];
//					print(x<<1|1,i+1);
//					printf(" ");
//					print(v,now);
//					printf("\n");
				}
//				printf("val=%d\n",f[0][0]);
				f[j][(x<<j)+y]=0;
			}
		}
	}
	for(int i=l-1;i>=0;i--)printf("%d",(ans>>i)&1);
	printf("\n");
	return 0;
}
### Prompt

In CPP, your task is to solve 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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,k,f[21][1<<20],ans,l;
char s[2000005];
void print(int x,int l)
{
	for(int i=l-1;i>=0;i--)
	  printf("%d",(x>>i)&1);
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=0;i<=n;i++)
	{
		scanf("%s",s);
		for(int j=0;j<(1<<i);j++)
		  if(s[j]=='1')f[i][j]=1;
	}
	for(int i=0;i<=n;i++)
	{
		for(int j=1;j<=n-i;j++)
		{
			for(int x=0;x<(1<<i);x++)
			for(int y=0;y<(1<<j);y++)
			{
				if(!f[j][(x<<j)+y])continue;
				f[0][x]+=f[j][(x<<j)+y];
//				print(x,i);
//				printf(" ");
//				print(y,j);
//				printf(" %d,val=%d\n",f[j][(x<<j)+y],f[0][0]);
			}
		}
		for(int x=(1<<i)-1;x>=0;x--)
		{
//				print(x,i);
//				printf(" %d\n",f[0][x]);
			if(f[0][x]>=k)ans=x,l=i; 
			f[0][x]=0;
		}
		for(int j=1;j<=n-i;j++)
		{
			for(int x=0;x<(1<<i);x++)
			for(int y=0;y<(1<<j);y++)
			{
				if(!f[j][(x<<j)+y])continue;
//				printf("f[");
//				print(x,i);
//				printf("][");
//				print(y,j);
//				printf("]=%d\n",f[j][(x<<j)+y]);
				int now=j-1,v=y;
				while(now>=0&&((1<<now)&y))v-=(1<<now),now--;
				if(now>=0)
				{
					f[now][(x<<(now+1))+v]+=f[j][(x<<j)+y];
//					print(x<<1,i+1);
//					printf(" ");
//					print(v,now);
//					printf("\n");
				}
				now=j-1,v=y;
				while(now>=0&&!((1<<now)&y))now--;
				if(now>=0)
				{
					f[now][(x<<(now+1))+v]+=f[j][(x<<j)+y];
//					print(x<<1|1,i+1);
//					printf(" ");
//					print(v,now);
//					printf("\n");
				}
//				printf("val=%d\n",f[0][0]);
				f[j][(x<<j)+y]=0;
			}
		}
	}
	for(int i=l-1;i>=0;i--)printf("%d",(ans>>i)&1);
	printf("\n");
	return 0;
}
```