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:
//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 unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;

/*  ------  by:duyi  ------  */ // myt天下第一
int n,K,dp[21][1<<21];
char s[(1<<20)+5];
int main() {
	scanf("%d%d",&n,&K);
	for(int i=0;i<=n;++i){
		scanf("%s",s);
		for(int j=0;j<(1<<i);++j)if(s[j]=='1'){
			dp[i][j|(1<<i)]=1;
		}
	}
	for(int i=n;i>=1;--i){//未确定部分的长度
		for(int j=i;j<=n;++j){//整个串的长度
			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-1;t>=0;--t)if(((k>>t)&1)==1){p=t;break;}
				for(int t=i-1;t>=0;--t)if(((k>>t)&1)==0){q=t;break;}
				dp[0][(k+b)>>i]+=tmp;
				if(p!=-1)dp[p][((((k>>i)<<1)|1)<<p)|(((1<<p)-1)&k)|(1<<(j-(i-p)+1))]+=tmp;
				if(q!=-1)dp[q][(((k>>i)<<1)<<q)|(((1<<q)-1)&k)|(1<<(j-(i-q)+1))]+=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);
}