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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#define LL long long
#define LD long double
using namespace std;
const int NN=2100000 +117;
const int MM=22;
int read(){
	int fl=1,x;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if(c=='-'){fl=-1;c=getchar();}
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<3)+(x<<1)+c-'0';
	return x*fl;
}
void open(){
	freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
}
void close(){
	fclose(stdin);
	fclose(stdout);
}


int m,n;
string s[MM]={};
int dp[MM][NN]={};
int res[NN]={};
int trans[NN][2]={};
int lg[NN]={};
char op[NN]={};
void gettrans(){
	for(int i=1;i<1<<(n+1);++i){
		for(int j=lg[i]-1;j>=0;--j){
			if(i&(1<<j)){
				trans[i][1]=i&((1<<(j+1))-1);
				trans[i][1]|=1<<(j+1);
				break;
			}
		}
		for(int j=lg[i]-1;j>=0;--j){
			if((i&(1<<j))==0){
				trans[i][0]=i&((1<<(j+1))-1);
				trans[i][0]|=1<<(j+1);
				break;
			}
		}
	}
}
int main(){
	//open();
	n=read();
	int k=read();
	for(int i=0;i<=n;++i){
		cin>>s[i];
		for(int j=0;j<(1<<i);++j){
			if(s[i][j]-'0')dp[0][j+(1<<i)]=1;
		}
	}
	int all=1<<(n+1);
	for(int i=0;i<=n;++i){
		lg[1<<i]=i;
	}
	for(int i=2;i<all;++i){
		if(!lg[i])lg[i]=lg[i-1];
	}
	gettrans();
	for(int i=0;i<=n;++i){
		for(int j=1<<(i);j<all;++j){
			int len=lg[j];
			int las=j&((1<<(len-i))-1);
			las|=1<<(len-i);
			int now=j>>(len-i);
			res[now]+=dp[i][j];
			//if(dp[i][j])cerr<<now<<" "<<las<<" "<<dp[i][j]<<endl;
			if(trans[las][0]){
				int xtl=lg[trans[las][0]];
				int xt=trans[las][0]-(1<<xtl);
				xt=xt|(now<<xtl);
				dp[i+1][xt]+=dp[i][j];
			}
			if(trans[las][1]){
				int xtl=lg[trans[las][1]];
				int xt=trans[las][1]-(1<<xtl);
				xt=xt|(now<<xtl);
				dp[i+1][xt]+=dp[i][j];
			}
		}
	}
	for(int len=n;len>=0;--len){
		for(int j=1<<len;j<1<<(len+1);++j){
			int now=j;
			if(res[now]>=k){
				for(int i=len-1;i>=0;--i){
					op[i]=now&1;
					now>>=1;
				}
				for(int i=0;i<len;++i){
					putchar(op[i]+'0');
				}
				putchar('\n');
				return 0;
			}
		}
	}
	printf("-1\n");
	return 0;
	close();
	return 0;
}