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>
using namespace std;

#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=20;
int n,k,tot;
int dp[1<<maxn+1][2][4];
int id[maxn+1][1<<maxn],cnt[1<<maxn+1];
char s[1<<maxn];

int main(){
	scanf("%d%d",&n,&k);
	rep(len,n+1)rep(mask,1<<len)id[len][mask]=tot++;
	rep(len,n+1){
		scanf("%s",s);
		rep(mask,1<<len)if(s[mask]=='1')
			dp[id[len][mask]][0][0]++;
	}
	rep(i,n+1)for(int len=n;len>=0;len--)rep(mask,1<<len)rep(f,4){
		int cur=dp[id[len][mask]][i&1][f];
		if(!cur)continue;
		if(i==len){
			cnt[id[len][mask]]+=cur;
			continue;
		}
		int c=((mask>>i)&1);
		if(!((f>>c)&1))dp[id[len][mask]][(i+1)&1][0]+=cur;
		if(len)dp[id[len-1][(mask&((1<<i)-1))|((mask>>(i+1))<<i)]][i&1][f|(1<<c)]+=cur;
		dp[id[len][mask]][i&1][f]=0;
	}
	for(int len=n;len>=0;len--)rep(mask,1<<len){
		if(cnt[id[len][mask]]>=k){
			string ans="";
			rep(i,len)ans+=char(((mask>>i)&1)+'0');
			reverse(ans.begin(),ans.end());
			printf("%s\n",ans.c_str());
			return 0;
		}
	}
	return 0;
}