#include<bits/stdc++.h>
using namespace std;
template<typename T>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
namespace run{
	const int N=1<<21;
	int n,K,tr[N][2],f[21][N],count[N];
	inline void print(int A){
		for(int i=count[A];~i;i--) cout<<(A>>i&1);
		cout<<endl;
	}
	int main(){
		n=read<int>(),K=read<int>();
		count[0]=-1;
		for(int i=1;i<2<<n;i++) count[i]=count[i>>1]+1;
		for(int i=0;i<=n;i++)
			for(int j=0;j<1<<i;j++) scanf("%1d",&f[i][1<<i^j]);
		tr[0][0]=tr[0][1]=tr[1][0]=tr[1][1]=0;
		for(int i=2;i<2<<n;i++){
			if(i>>(count[i]-1)&1){
				tr[i][1]=i^1<<count[i];
				tr[i][0]=tr[tr[i][1]][0];
			}else{
				tr[i][0]=i^3<<(count[i]-1);
				tr[i][1]=tr[tr[i][0]][1];
			}
		}

		for(int j=n;j;j--)
			for(int i=2;i<2<<n;i++) if(f[j][i]){
				int A=i>>j,B=(i&((1<<j)-1))^(1<<j);
				if(tr[B][0]){
					int tmp=tr[B][0],len=count[tmp];
					f[len][A<<(len+1)^(tmp^(1<<len))]+=f[j][i];
				}
				if(tr[B][1]){
					int tmp=tr[B][1],len=count[tmp];
					f[len][((A<<1)+1)<<len^(tmp^(1<<len))]+=f[j][i];
				}
				f[0][A]+=f[j][i];
			}
		int ans=1;
		for(int i=2;i<2<<n;i++) if(f[0][i]>=K && count[i]>count[ans]) ans=i;
		if(ans>1)
			for(int j=count[ans]-1;~j;j--) printf("%d",ans>>j&1);
		return 0;
	}
}
int main(){
#ifdef my
	freopen("subsequence.in","r",stdin);
	freopen("subsequence.out","w",stdout);
#endif
	return run::main();
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;
template<typename T>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
namespace run{
	const int N=1<<21;
	int n,K,tr[N][2],f[21][N],count[N];
	inline void print(int A){
		for(int i=count[A];~i;i--) cout<<(A>>i&1);
		cout<<endl;
	}
	int main(){
		n=read<int>(),K=read<int>();
		count[0]=-1;
		for(int i=1;i<2<<n;i++) count[i]=count[i>>1]+1;
		for(int i=0;i<=n;i++)
			for(int j=0;j<1<<i;j++) scanf("%1d",&f[i][1<<i^j]);
		tr[0][0]=tr[0][1]=tr[1][0]=tr[1][1]=0;
		for(int i=2;i<2<<n;i++){
			if(i>>(count[i]-1)&1){
				tr[i][1]=i^1<<count[i];
				tr[i][0]=tr[tr[i][1]][0];
			}else{
				tr[i][0]=i^3<<(count[i]-1);
				tr[i][1]=tr[tr[i][0]][1];
			}
		}

		for(int j=n;j;j--)
			for(int i=2;i<2<<n;i++) if(f[j][i]){
				int A=i>>j,B=(i&((1<<j)-1))^(1<<j);
				if(tr[B][0]){
					int tmp=tr[B][0],len=count[tmp];
					f[len][A<<(len+1)^(tmp^(1<<len))]+=f[j][i];
				}
				if(tr[B][1]){
					int tmp=tr[B][1],len=count[tmp];
					f[len][((A<<1)+1)<<len^(tmp^(1<<len))]+=f[j][i];
				}
				f[0][A]+=f[j][i];
			}
		int ans=1;
		for(int i=2;i<2<<n;i++) if(f[0][i]>=K && count[i]>count[ans]) ans=i;
		if(ans>1)
			for(int j=count[ans]-1;~j;j--) printf("%d",ans>>j&1);
		return 0;
	}
}
int main(){
#ifdef my
	freopen("subsequence.in","r",stdin);
	freopen("subsequence.out","w",stdout);
#endif
	return run::main();
}
```