#include<vector>
#include<cmath>
#include<map>
#include<cstdlib>
#include<iostream>
#include<sstream>
#include<fstream>
#include<string>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<set>
#include<stack>
#include<bitset>
#include<functional>
#include<ctime>
#include<queue>
#include<deque>
#include<complex>
#include<cassert>
using namespace std;
#define pb push_back
#define pf push_front
typedef long long lint;
typedef complex<double> P;
#define mp make_pair
#define fi first
#define se second
typedef pair<int,int> pint;
#define All(s) s.begin(),s.end()
#define rAll(s) s.rbegin(),s.rend()
#define REP(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)
//問題文および制約はちゃんと確認しよう！
//サイズは10^5じゃなくて2×10^5とかかもしれないし、重要な制約・条件を見落としているかも
//とりあえずサンプルを読んでから解法を考えよう？
int dp[(1<<21)+10][22];
int main()
{
	memset(dp,0,sizeof(dp));
	int n,K;cin>>n>>K;string s;
	rep(i,n+1){
		cin>>s;
		rep(j,(1<<i)){
			if(s[j]=='1') dp[(1<<i)+j][i]++;
		}
	}
	for(int i=n;i>0;i--) rep(j,(1<<i)) rep(l,i+1){
		if(dp[(1<<i)+j][l]<1) continue;
		//cout<<i<<' '<<j<<' '<<l<<' '<<dp[(1<<i)+j][l]<<endl;
		rep(k,l){
			if(k+1<i && ((j&(1<<k))>>k)==((j&(1<<(k+1)))>>(k+1))) continue;
			int to=(j>>(k+1)),bo=(j&((1<<k)-1));
			//cout<<j<<' '<<k<<' '<<(to<<k)+bo<<endl;
			//if(i==4 && (to<<k)+bo==4) cout<<j<<' '<<k<<endl;
			if(i==3 && j==1){
				//cout<<(to<<k)+bo<<' '<<k<<endl;
			}
			dp[(1<<(i-1))+(to<<k)+bo][k]+=dp[(1<<i)+j][l];
		}
		//cout<<dp[4][2]<<endl;
	}
	for(int i=n;i>0;i--) rep(j,(1<<i)){
		int sum=0;
		rep(k,i+1) sum+=dp[(1<<i)+j][k];
		if(sum>=K){
			string out="";
			rep(l,i){
				out+=('0'+(j%2));j/=2;
			}
			reverse(All(out));
			cout<<out<<endl;return 0;
		}
	}
	cout<<""<<endl;
}

### 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<vector>
#include<cmath>
#include<map>
#include<cstdlib>
#include<iostream>
#include<sstream>
#include<fstream>
#include<string>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<set>
#include<stack>
#include<bitset>
#include<functional>
#include<ctime>
#include<queue>
#include<deque>
#include<complex>
#include<cassert>
using namespace std;
#define pb push_back
#define pf push_front
typedef long long lint;
typedef complex<double> P;
#define mp make_pair
#define fi first
#define se second
typedef pair<int,int> pint;
#define All(s) s.begin(),s.end()
#define rAll(s) s.rbegin(),s.rend()
#define REP(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)
//問題文および制約はちゃんと確認しよう！
//サイズは10^5じゃなくて2×10^5とかかもしれないし、重要な制約・条件を見落としているかも
//とりあえずサンプルを読んでから解法を考えよう？
int dp[(1<<21)+10][22];
int main()
{
	memset(dp,0,sizeof(dp));
	int n,K;cin>>n>>K;string s;
	rep(i,n+1){
		cin>>s;
		rep(j,(1<<i)){
			if(s[j]=='1') dp[(1<<i)+j][i]++;
		}
	}
	for(int i=n;i>0;i--) rep(j,(1<<i)) rep(l,i+1){
		if(dp[(1<<i)+j][l]<1) continue;
		//cout<<i<<' '<<j<<' '<<l<<' '<<dp[(1<<i)+j][l]<<endl;
		rep(k,l){
			if(k+1<i && ((j&(1<<k))>>k)==((j&(1<<(k+1)))>>(k+1))) continue;
			int to=(j>>(k+1)),bo=(j&((1<<k)-1));
			//cout<<j<<' '<<k<<' '<<(to<<k)+bo<<endl;
			//if(i==4 && (to<<k)+bo==4) cout<<j<<' '<<k<<endl;
			if(i==3 && j==1){
				//cout<<(to<<k)+bo<<' '<<k<<endl;
			}
			dp[(1<<(i-1))+(to<<k)+bo][k]+=dp[(1<<i)+j][l];
		}
		//cout<<dp[4][2]<<endl;
	}
	for(int i=n;i>0;i--) rep(j,(1<<i)){
		int sum=0;
		rep(k,i+1) sum+=dp[(1<<i)+j][k];
		if(sum>=K){
			string out="";
			rep(l,i){
				out+=('0'+(j%2));j/=2;
			}
			reverse(All(out));
			cout<<out<<endl;return 0;
		}
	}
	cout<<""<<endl;
}

```