//problem:
#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define mk make_pair
#define lob lower_bound
#define upb upper_bound
#define fst first
#define scd second

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

inline int read(){
	int f=1,x=0;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline ll readll(){
	ll f=1,x=0;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

char s[1<<21];
int n,K,dp[25][1<<21];
int main() {
	scanf("%d%d",&n,&K);
	for(int i=0;i<=n;++i){
		scanf("%s",s);
		for(int j=0;j<=(1<<i)-1;++j){
			if(s[j]=='1')dp[i][j|(1<<i)]=1;
		}
	}
	for(int i=n;i>=1;--i){
		for(int j=i;j<=n;++j){
			int b=(1<<j);
			for(int k=0;k<b;++k){
				int tmp=dp[i][k|b],p=-1,q=-1;if(!tmp)continue;
				for(int t=i;t>=1;--t)if(((k>>(t-1))&1)==1){p=t;break;}
				for(int t=i;t>=1;--t)if(((k>>(t-1))&1)==0){q=t;break;}
				dp[0][(k+b)>>i]+=tmp;
				if(p!=-1)dp[p-1][(k&((1<<(p-1))-1))|(((k>>i)<<1|1)<<(p-1))|(b>>(i-p))]+=tmp;
				if(q!=-1)dp[q-1][(k&((1<<(q-1))-1))|(((k>>i)<<1)<<(q-1))|(b>>(i-q))]+=tmp;
			}
		}
	}
	for(int i=n;i>=0;--i){
		for(int j=0;j<(1<<i);++j){
			if(dp[0][j|(1<<i)]>=K){
				for(int k=i-1;k>=0;--k)if((j>>k)&1)putchar('1');else putchar('0');
				puts("");
				return 0;
			}
		}
	}
	assert(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
//problem:
#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define mk make_pair
#define lob lower_bound
#define upb upper_bound
#define fst first
#define scd second

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

inline int read(){
	int f=1,x=0;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline ll readll(){
	ll f=1,x=0;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

char s[1<<21];
int n,K,dp[25][1<<21];
int main() {
	scanf("%d%d",&n,&K);
	for(int i=0;i<=n;++i){
		scanf("%s",s);
		for(int j=0;j<=(1<<i)-1;++j){
			if(s[j]=='1')dp[i][j|(1<<i)]=1;
		}
	}
	for(int i=n;i>=1;--i){
		for(int j=i;j<=n;++j){
			int b=(1<<j);
			for(int k=0;k<b;++k){
				int tmp=dp[i][k|b],p=-1,q=-1;if(!tmp)continue;
				for(int t=i;t>=1;--t)if(((k>>(t-1))&1)==1){p=t;break;}
				for(int t=i;t>=1;--t)if(((k>>(t-1))&1)==0){q=t;break;}
				dp[0][(k+b)>>i]+=tmp;
				if(p!=-1)dp[p-1][(k&((1<<(p-1))-1))|(((k>>i)<<1|1)<<(p-1))|(b>>(i-p))]+=tmp;
				if(q!=-1)dp[q-1][(k&((1<<(q-1))-1))|(((k>>i)<<1)<<(q-1))|(b>>(i-q))]+=tmp;
			}
		}
	}
	for(int i=n;i>=0;--i){
		for(int j=0;j<(1<<i);++j){
			if(dp[0][j|(1<<i)]>=K){
				for(int k=i-1;k>=0;--k)if((j>>k)&1)putchar('1');else putchar('0');
				puts("");
				return 0;
			}
		}
	}
	assert(0);
	return 0;
}
```