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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,K;
int f[2100000][22];
char s[2100000];
int main() {
	scanf("%d%d",&n,&K);
	for(int i=0;i<=n;i++){
		scanf("%s",s);
		int lim=(1<<i);
		for (int j=0;j<lim;j++) f[lim|j][i]=(s[j]-'0');
	}
	for (int i=n;i>=0;i--)
		for (int s=0;s<(1<<i);s++)
			for (int j=i;j>=0;j--){
				int cur=f[(1<<i)|s][j];
				if (cur){
					int s1=s>>j;
					int s2=s&((1<<j)-1);
					if (j) f[(1<<(i-j))|s1][0]+=cur;
					if (s2){
						int p=j-1;
						while (!((s2>>p)&1)) p--;
						f[(1<<(i-(j-1-p)))|(s1<<(p+1))|(s2&((1<<p)-1))|(1<<p)][p]+=cur;
					}
					if (((1<<j)-1)&(~s2)){
						int p=j-1;
						while ((s2>>p)&1) p--;
						f[(1<<(i-(j-1-p)))|(s1<<(p+1))|(s2&((1<<p)-1))][p]+=cur;
					}
				}
			}
	for (int i=n;i>=0;i--)
		for (int j=0;j<(1<<i);j++)
			if (f[(1<<i)|j][0]>=K){
				for(int k=i-1;k>=0;k--) printf("%d",(j>>k)&1);
				return 0;
			}
	return 0;
}