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 M 21
#define N (1<<M)

int n,m,a[M][N],H[N],f[M][N],g[M][N];
char s[N];

int main(){
	scanf("%d%d",&n,&m);
	for (int i=0;i<=n;++i){
		scanf("%s",s);
		for (int j=0;j<1<<i;++j) a[i][j]=s[j];
	}
	for (int i=1;i<1<<(n+1);++i)
		for (int j=0;j<=n;++j)
			if (i>>j&1) H[i]=j;
	for (int i=(1<<(n+1))-1;i;--i){
		int t=H[i];
		f[0][i]=a[t][i^(1<<t)]-'0';
		if (!(i>>n&1)) f[0][i]+=f[0][i<<1]+f[0][i<<1|1];
	}
	for (int i=1;i<=n;++i){
		for (int j=1;j<1<<(n+1);++j){
			int t=H[j];
			if (t>=i){
				int x=j;
				int y=x%(1<<i);
				x>>=i;
				int o=y>>(i-1)&1;
				for (int k=1;t+k-1<=n;++k){
					x<<=1; x|=o;
					f[i][j]+=f[i-1][(x<<(i-1))|(y^(o<<(i-1)))];
					x^=1;
				}
				if (t==i) g[i][y]+=f[i][j];
			}
		}
	}
	for (int i=n;i;--i)
		for (int j=0;j<1<<n;++j)
			if (g[i][j]>=m){
				for (int k=i-1;~k;--k) printf("%d",j>>k&1);
				puts("");
				return 0;
			}
	
	return 0;
}