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

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

template<class A,class B>
ostream& operator<<(ostream& ost,const pair<A,B>&p){
	ost<<"{"<<p.first<<","<<p.second<<"}";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost,const vector<T>&v){
	ost<<"{";
	for(int i=0;i<v.size();i++){
		if(i)ost<<",";
		ost<<v[i];
	}
	ost<<"}";
	return ost;
}

int T[1<<21],B[1<<21];

int dp[1<<21];
int ans[1<<21];

signed main(){
	for(int i=1;i<1<<21;i++){
		B[i]=__builtin_ctz(i);
		T[i]=31-__builtin_clz(i);
	}
	int N,K;
	cin>>N>>K;


	rep(i,N+1){
		string S;cin>>S;
		rep(j,1<<i){
			if(S[j]=='0')continue;
			ans[1]++;
			dp[j|1<<i]++;
		}
	}

	for(int i=1;i<=N;i++){
		for(int j=1;j<1<<N+1;j++){
			int tmp=dp[j];
			dp[j]=0;

			int t=T[j];
			if(t<i)continue;
			int cur=j&((1<<i-1)-1);
			int b=j-cur;
			
			rep(uku,2){
				int k;
				if(uku==0)k=B[b];
				else k=B[(~(b>>(i-1)))&((1<<21)-1)]+i-1;
				if(k>=t)continue;
				int bb=cur|(b>>k<<(i-1));
				dp[bb]+=tmp;
			}
		}
		for(int j=1;j<1<<N+1;j++){
			int b=j&((1<<i)-1);
			b+=1<<i;
			ans[b]+=dp[j];
		}
	}


	int len=-1;
	for(int j=1;j<1<<N+1;j++)if(ans[j]>=K)len=T[j];

	for(int j=1<<len;j<1<<len+1;j++)if(ans[j]>=K){
		rep(k,len)cout<<(j>>(len-1-k)&1);
		cout<<endl;
		return 0;
	}
	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 int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

template<class A,class B>
ostream& operator<<(ostream& ost,const pair<A,B>&p){
	ost<<"{"<<p.first<<","<<p.second<<"}";
	return ost;
}

template<class T>
ostream& operator<<(ostream& ost,const vector<T>&v){
	ost<<"{";
	for(int i=0;i<v.size();i++){
		if(i)ost<<",";
		ost<<v[i];
	}
	ost<<"}";
	return ost;
}

int T[1<<21],B[1<<21];

int dp[1<<21];
int ans[1<<21];

signed main(){
	for(int i=1;i<1<<21;i++){
		B[i]=__builtin_ctz(i);
		T[i]=31-__builtin_clz(i);
	}
	int N,K;
	cin>>N>>K;


	rep(i,N+1){
		string S;cin>>S;
		rep(j,1<<i){
			if(S[j]=='0')continue;
			ans[1]++;
			dp[j|1<<i]++;
		}
	}

	for(int i=1;i<=N;i++){
		for(int j=1;j<1<<N+1;j++){
			int tmp=dp[j];
			dp[j]=0;

			int t=T[j];
			if(t<i)continue;
			int cur=j&((1<<i-1)-1);
			int b=j-cur;
			
			rep(uku,2){
				int k;
				if(uku==0)k=B[b];
				else k=B[(~(b>>(i-1)))&((1<<21)-1)]+i-1;
				if(k>=t)continue;
				int bb=cur|(b>>k<<(i-1));
				dp[bb]+=tmp;
			}
		}
		for(int j=1;j<1<<N+1;j++){
			int b=j&((1<<i)-1);
			b+=1<<i;
			ans[b]+=dp[j];
		}
	}


	int len=-1;
	for(int j=1;j<1<<N+1;j++)if(ans[j]>=K)len=T[j];

	for(int j=1<<len;j<1<<len+1;j++)if(ans[j]>=K){
		rep(k,len)cout<<(j>>(len-1-k)&1);
		cout<<endl;
		return 0;
	}
	return 0;
}
```