#include<cstdio>
#include<cstring>
using namespace std;
const int N=22,M=1e6+1e5;
int f[N][2*M],d[N],g[M],Ans[N],n,K;char s[M];bool bk[N][M];
int main()
{
	scanf("%d%d",&n,&K);
	d[0]=1;for(int i=1;i<=n+1;i++) d[i]=d[i-1]*2;
	for(int i=0;i<=n;i++){
		scanf("%s",s+1);
		for(int j=0;j<d[i];j++)
			if(s[j+1]=='1') bk[i][j]=true;
	}
	memset(f,0,sizeof(f));
	for(int i=1;i<=n;i++)
		for(int j=d[i]-1;j<d[i+1]-1;j++)
			if(bk[i][j-d[i]+1]) f[i][j]=1;
	for(int i=n;i>=1;i--)
		for(int j=i;j>0;j--)
			for(int k=0;k<=d[i]-1;k++){
				int t=k%d[j],t0=k/d[j],t1=0,t2=0;
				for(int l=j;l>=1;l--)
					if(k/d[l-1]%2==0) {t1=l;break;}
				for(int l=j;l>=1;l--)
					if(k/d[l-1]%2==1) {t2=l;break;}
				if(t1){
					int t3=(t0*2)*d[t1-1]+k%d[t1-1];
					f[t1-1][t3+d[i-j+t1]-1]+=f[j][k+d[i]-1];
				}
				if(t2){
					int t3=(t0*2+1)*d[t2-1]+k%d[t2-1];
					f[t2-1][t3+d[i-j+t2]-1]+=f[j][k+d[i]-1];
				}
			}
	for(int i=n;i>=1;i--)
		for(int j=0;j<d[i];j++){
			int Sum=0;
			for(int k=0;k<=n-i;k++)
				for(int l=0;l<d[k];l++){
					int t=j*d[k]+l+d[i+k]-1;
					Sum+=f[k][t];
				}
			if(Sum>=K){
				for(int p=1;p<=i;p++) Ans[p]=j%2,j/=2;
				for(int p=i;p>=1;p--) printf("%d",Ans[p]);
				printf("\n");
				return 0;
			}
		}
	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
#include<cstdio>
#include<cstring>
using namespace std;
const int N=22,M=1e6+1e5;
int f[N][2*M],d[N],g[M],Ans[N],n,K;char s[M];bool bk[N][M];
int main()
{
	scanf("%d%d",&n,&K);
	d[0]=1;for(int i=1;i<=n+1;i++) d[i]=d[i-1]*2;
	for(int i=0;i<=n;i++){
		scanf("%s",s+1);
		for(int j=0;j<d[i];j++)
			if(s[j+1]=='1') bk[i][j]=true;
	}
	memset(f,0,sizeof(f));
	for(int i=1;i<=n;i++)
		for(int j=d[i]-1;j<d[i+1]-1;j++)
			if(bk[i][j-d[i]+1]) f[i][j]=1;
	for(int i=n;i>=1;i--)
		for(int j=i;j>0;j--)
			for(int k=0;k<=d[i]-1;k++){
				int t=k%d[j],t0=k/d[j],t1=0,t2=0;
				for(int l=j;l>=1;l--)
					if(k/d[l-1]%2==0) {t1=l;break;}
				for(int l=j;l>=1;l--)
					if(k/d[l-1]%2==1) {t2=l;break;}
				if(t1){
					int t3=(t0*2)*d[t1-1]+k%d[t1-1];
					f[t1-1][t3+d[i-j+t1]-1]+=f[j][k+d[i]-1];
				}
				if(t2){
					int t3=(t0*2+1)*d[t2-1]+k%d[t2-1];
					f[t2-1][t3+d[i-j+t2]-1]+=f[j][k+d[i]-1];
				}
			}
	for(int i=n;i>=1;i--)
		for(int j=0;j<d[i];j++){
			int Sum=0;
			for(int k=0;k<=n-i;k++)
				for(int l=0;l<d[k];l++){
					int t=j*d[k]+l+d[i+k]-1;
					Sum+=f[k][t];
				}
			if(Sum>=K){
				for(int p=1;p<=i;p++) Ans[p]=j%2,j/=2;
				for(int p=i;p>=1;p--) printf("%d",Ans[p]);
				printf("\n");
				return 0;
			}
		}
	return 0;
}
```