#pragma GCC optimize("unroll-loops")
#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
char s[21][(1<<20)+5];
int Arr[41943040+5],*tl=Arr;
int *sm[25][25],w0[1<<20][20],w1[1<<20][20];
int main()
{
	int n,kk;
	scanf("%d%d",&n,&kk);
	for(int i=0;i<=n;i++)
		scanf("%s",s[i]);
	for(int i=0;i<(1<<n);i++){
		if(i&1)w1[i][0]=0,w0[i][0]=-1;
		else w1[i][0]=-1,w0[i][0]=0;
		for(int j=1;j<n;j++)
			if(i>>j&1)w1[i][j]=j,w0[i][j]=w0[i][j-1];
			else w1[i][j]=w1[i][j-1],w0[i][j]=j;
	}
	for(int i=1;i<=n;i++)
		for(int j=i;j>=0;--j)
			sm[j][i]=tl,tl+=1<<i;
	cerr<<tl-Arr<<" >>\n";
	int len=0,as=0,all=(1<<20)-1;
	for(int p=n;p;--p){
		for(int i=0;i<(1<<p);i++)
			if(s[p][i]=='1')
				++sm[p][p][i];for(int i=0;i<(1<<p);i++)
		for(int j=p;j;--j)
			{
				if(j<p)sm[0][p-j][i>>j]+=sm[j][p][i];
				if(w1[i][j-1]!=-1){
					int c=w1[i][j-1];
					sm[c][p-(j-1-c)]
					[((i&(all^((1<<j)-1)))>>(j-1-c))|(i&((1<<(c+1))-1))]
					+=sm[j][p][i];
				}
				if(w0[i][j-1]!=-1){
					int c=w0[i][j-1];
					sm[c][p-(j-1-c)]
					[((i&(all^((1<<j)-1)))>>(j-1-c))|(i&((1<<(c+1))-1))]
					+=sm[j][p][i];
				}
			}
		for(int i=0;i<(1<<p);i++)
			if(sm[0][p][i]>=kk){
				len=p;
				as=i;
				break;
			}
		if(len)break;
	}
	if(!len)return 0&puts("");
	for(int i=len-1;i>=0;--i)
		printf("%d",as>>i&1);
	return 0;
}
### Prompt

Create a solution in CPP for 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
#pragma GCC optimize("unroll-loops")
#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
char s[21][(1<<20)+5];
int Arr[41943040+5],*tl=Arr;
int *sm[25][25],w0[1<<20][20],w1[1<<20][20];
int main()
{
	int n,kk;
	scanf("%d%d",&n,&kk);
	for(int i=0;i<=n;i++)
		scanf("%s",s[i]);
	for(int i=0;i<(1<<n);i++){
		if(i&1)w1[i][0]=0,w0[i][0]=-1;
		else w1[i][0]=-1,w0[i][0]=0;
		for(int j=1;j<n;j++)
			if(i>>j&1)w1[i][j]=j,w0[i][j]=w0[i][j-1];
			else w1[i][j]=w1[i][j-1],w0[i][j]=j;
	}
	for(int i=1;i<=n;i++)
		for(int j=i;j>=0;--j)
			sm[j][i]=tl,tl+=1<<i;
	cerr<<tl-Arr<<" >>\n";
	int len=0,as=0,all=(1<<20)-1;
	for(int p=n;p;--p){
		for(int i=0;i<(1<<p);i++)
			if(s[p][i]=='1')
				++sm[p][p][i];for(int i=0;i<(1<<p);i++)
		for(int j=p;j;--j)
			{
				if(j<p)sm[0][p-j][i>>j]+=sm[j][p][i];
				if(w1[i][j-1]!=-1){
					int c=w1[i][j-1];
					sm[c][p-(j-1-c)]
					[((i&(all^((1<<j)-1)))>>(j-1-c))|(i&((1<<(c+1))-1))]
					+=sm[j][p][i];
				}
				if(w0[i][j-1]!=-1){
					int c=w0[i][j-1];
					sm[c][p-(j-1-c)]
					[((i&(all^((1<<j)-1)))>>(j-1-c))|(i&((1<<(c+1))-1))]
					+=sm[j][p][i];
				}
			}
		for(int i=0;i<(1<<p);i++)
			if(sm[0][p][i]>=kk){
				len=p;
				as=i;
				break;
			}
		if(len)break;
	}
	if(!len)return 0&puts("");
	for(int i=len-1;i>=0;--i)
		printf("%d",as>>i&1);
	return 0;
}
```