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>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=24,S=1200120;
int n,K,L,ans,f[N][S],g[N][S];
char s[S];
int main(){
	scanf("%d%d",&n,&K);
	FOR(i,0,n){
		scanf("%s",s);
		FOR(j,0,(1<<i)-1) f[i][j]=s[j]=='1'; 
	}
	FOR(i,1,n)FOR(j,0,(1<<i)-1){
		int t=j>>i-1&1,r=i-1;
		for(;r && (j>>r-1&1)==t;r--);g[i][j]=r;
	}
	FOR(i,0,n){
		FOR(j,1,n-i)FOR(k,0,(1<<i+j)-1){
			f[i][k>>j]+=f[i+j][k];
			int t=g[j][k&(1<<j)-1];
			if(t) f[i+t][(k>>j<<t)|(k&(1<<t)-1)]+=f[i+j][k];
		}
		ROF(j,(1<<i)-1,0)if(f[i][j]>=K) L=i,ans=j;
	}
	for(L--;~L;L--) cout<<(ans>>L&1);cout<<"\n";
}/*
5 100
0
01
1011
11010111
0001110111011110
01100100100011110101111111011010
*/