#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,lenmax=0,K,Ans=0;
#define V 2300010
int *tree[V];int ans[V];
int to[V][2];
char str[1<<20];
void Add(int &a,int b){a+=b;}
int main(){
	scanf("%d%d",&n,&K);
	for(int i=2;i<(1<<(n+1));++i){
		if(to[i>>1][0])to[i][0]=to[i>>1][0]*2+(i&1);
		else if(!(i&1))to[i][0]=1;
		if(to[i>>1][1])to[i][1]=to[i>>1][1]*2+(i&1);
		else if(i&1)to[i][1]=1;
	}
	tree[1]=new int[(1<<(n+1))];
	tree[1][0]=1;
	for(int i=0;i<=n;++i){
		scanf("%s",str);
		for(int j=0;j<(1<<i);++j)
			if(str[j]=='1')tree[1][j|(1<<i)]=1;
			else tree[1][j|(1<<i)]=0;
	}
	for(int i=0;i<=n;++i){
		if(i<n)
			for(int j=0;j<(1<<(i+1));++j){
				tree[j|(1<<(i+1))]=new int[1<<(n-i)];
			    memset(tree[j|(1<<(i+1))],0,sizeof(int)*(1<<(n-i)));
			}
		for(int j=(1<<i);j<(1<<(i+1));++j){
			for(int k=1;k<(1<<(n-i+1));++k){
				if(i<n&&to[k][0])Add(tree[j*2][to[k][0]],tree[j][k]);
				if(i<n&&to[k][1])Add(tree[j*2+1][to[k][1]],tree[j][k]);
				Add(ans[j],tree[j][k]);
			}
			if(ans[j]>=K&&lenmax<i)lenmax=i,Ans=j;
		}
	}
	for(int i=lenmax-1;i>=0;--i)
		if(Ans&(1<<i))printf("1");
		else printf("0");
	return 0;
}
### Prompt

Please formulate 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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,lenmax=0,K,Ans=0;
#define V 2300010
int *tree[V];int ans[V];
int to[V][2];
char str[1<<20];
void Add(int &a,int b){a+=b;}
int main(){
	scanf("%d%d",&n,&K);
	for(int i=2;i<(1<<(n+1));++i){
		if(to[i>>1][0])to[i][0]=to[i>>1][0]*2+(i&1);
		else if(!(i&1))to[i][0]=1;
		if(to[i>>1][1])to[i][1]=to[i>>1][1]*2+(i&1);
		else if(i&1)to[i][1]=1;
	}
	tree[1]=new int[(1<<(n+1))];
	tree[1][0]=1;
	for(int i=0;i<=n;++i){
		scanf("%s",str);
		for(int j=0;j<(1<<i);++j)
			if(str[j]=='1')tree[1][j|(1<<i)]=1;
			else tree[1][j|(1<<i)]=0;
	}
	for(int i=0;i<=n;++i){
		if(i<n)
			for(int j=0;j<(1<<(i+1));++j){
				tree[j|(1<<(i+1))]=new int[1<<(n-i)];
			    memset(tree[j|(1<<(i+1))],0,sizeof(int)*(1<<(n-i)));
			}
		for(int j=(1<<i);j<(1<<(i+1));++j){
			for(int k=1;k<(1<<(n-i+1));++k){
				if(i<n&&to[k][0])Add(tree[j*2][to[k][0]],tree[j][k]);
				if(i<n&&to[k][1])Add(tree[j*2+1][to[k][1]],tree[j][k]);
				Add(ans[j],tree[j][k]);
			}
			if(ans[j]>=K&&lenmax<i)lenmax=i,Ans=j;
		}
	}
	for(int i=lenmax-1;i>=0;--i)
		if(Ans&(1<<i))printf("1");
		else printf("0");
	return 0;
}
```