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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=21,M=(1<<21)+5;
int trans[M][2],n,aim,len,best;
char st[M];
vi f[N];

void dfs(int k,int now){
	int sum=0;
	for(int i=1;i<f[k].size();i++)
		sum+=f[k][i];
	if(sum>=aim){
		if(k>len||k==len&&now<best)
			len=k,best=now;
	}
	if(k==n)return;
	REP(t,2){
		fill(f[k+1].begin(),f[k+1].end(),0);
		REP(i,f[k].size())
			if(f[k][i])f[k+1][trans[i][t]]+=f[k][i];
		dfs(k+1,(now<<1)|t);
	}
}

int main(){
	read(n),read(aim);
	rep(k,0,n){
		f[k].resize(1<<(n-k+1));
	}
	REP(i,1<<(n+1)){
		int s=i,k=n+1;
		while(k&&!(s>>k&1))k--;
		for(k--;k>=0;k--){
			if(!trans[i][(s>>k)&1])
				trans[i][(s>>k)&1]=(s&((1<<k)-1))|(1<<k);
		}
	}
	rep(k,0,n){
		scanf("%s",st);
		REP(i,1<<k)
			if(st[i]=='1')
				f[0][i|(1<<k)]=1;
	}
	dfs(0,1);
	per(i,len-1,0)
		putchar('0'+(best>>i&1));
	return puts(""),0;
}