#include<bits/stdc++.h>
using namespace std;
#define nn 20


int dp[2][1<<nn][nn+1][4];

void add(int &x,int y){
	x+=y;if(x>1e8) x=1e8;
}
/*
0->last ignore 0
1->last ignore 1
2->nothing
*/

int leng,bst;
int n,now;
void solve(int b){
	now^=1;
	if(b+1<=n) for(int i=0;i<(1<<b+1);i++){
		for(int j=0;j<=b;j++){
			int x=(i&(1<<j));int z=x>0;
			
			int o=i&((1<<j)-1);
			int e=(i-o-x)>>1;e|=o;
		//	cout<<e<<" "<<i<<" "<<j<<"\n";
		//	if(e<0) exit(0);
			for(int k=0;k<4;k++)
				add(dp[now^1][e][j][(1<<z)|k],dp[now][i][j][k]);
		}
	}
	now^=1;
	
	for(int i=0;i<(1<<b);i++){
		for(int j=0;j<b;j++){
			int x=(i&(1<<j));int z=x>0;
			
			for(int k=0;k<4;k++) if((k&(1<<z)));else 
				add(dp[now][i][j+1][0],dp[now][i][j][k]);
		}
	}
}

char str[22][1<<nn+1];int limit;
int main(){
	scanf("%d%d%s",&n,&limit,str[0]);assert(limit<=1e7);
	for(int i=1;i<=n;i++) scanf("%s",str[i]);
	
	for(int i=n;i>=1;i--){
		now^=1;
		for(int j=0;j<(1<<i);j++) memset(dp[now][j],0,sizeof dp[now][j]);
		
		for(int j=0;j<(1<<i);j++) if(str[i][j]=='1')
			add(dp[now][j][0][0],1);
			
		solve(i);bst=1e7;
		for(int j=0;j<(1<<i);j++) 
			if(dp[now][j][i][3]+dp[now][j][i][0]+dp[now][j][i][1]>=limit-dp[now][j][i][2])
				leng=i,bst=min(bst,j);
		
		if(leng){
			string g;
			while(bst){
				if(bst&1) g+='1';
				else g+='0';bst/=2;
				leng--;
			}
			while(leng--) g+='0';
			for(int i=g.size()-1;i>=0;i--)
				putchar(g[i]);
			exit(0);
		}
	}
	
	puts("");
	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<bits/stdc++.h>
using namespace std;
#define nn 20


int dp[2][1<<nn][nn+1][4];

void add(int &x,int y){
	x+=y;if(x>1e8) x=1e8;
}
/*
0->last ignore 0
1->last ignore 1
2->nothing
*/

int leng,bst;
int n,now;
void solve(int b){
	now^=1;
	if(b+1<=n) for(int i=0;i<(1<<b+1);i++){
		for(int j=0;j<=b;j++){
			int x=(i&(1<<j));int z=x>0;
			
			int o=i&((1<<j)-1);
			int e=(i-o-x)>>1;e|=o;
		//	cout<<e<<" "<<i<<" "<<j<<"\n";
		//	if(e<0) exit(0);
			for(int k=0;k<4;k++)
				add(dp[now^1][e][j][(1<<z)|k],dp[now][i][j][k]);
		}
	}
	now^=1;
	
	for(int i=0;i<(1<<b);i++){
		for(int j=0;j<b;j++){
			int x=(i&(1<<j));int z=x>0;
			
			for(int k=0;k<4;k++) if((k&(1<<z)));else 
				add(dp[now][i][j+1][0],dp[now][i][j][k]);
		}
	}
}

char str[22][1<<nn+1];int limit;
int main(){
	scanf("%d%d%s",&n,&limit,str[0]);assert(limit<=1e7);
	for(int i=1;i<=n;i++) scanf("%s",str[i]);
	
	for(int i=n;i>=1;i--){
		now^=1;
		for(int j=0;j<(1<<i);j++) memset(dp[now][j],0,sizeof dp[now][j]);
		
		for(int j=0;j<(1<<i);j++) if(str[i][j]=='1')
			add(dp[now][j][0][0],1);
			
		solve(i);bst=1e7;
		for(int j=0;j<(1<<i);j++) 
			if(dp[now][j][i][3]+dp[now][j][i][0]+dp[now][j][i][1]>=limit-dp[now][j][i][2])
				leng=i,bst=min(bst,j);
		
		if(leng){
			string g;
			while(bst){
				if(bst&1) g+='1';
				else g+='0';bst/=2;
				leng--;
			}
			while(leng--) g+='0';
			for(int i=g.size()-1;i>=0;i--)
				putchar(g[i]);
			exit(0);
		}
	}
	
	puts("");
	return 0;
}
```