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 <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
int f[25][(1<<20)+10];
int g[25][(1<<20)+10];
int n, m;
char s[(1<<20)+10];
int main () {
	int i, j, k;
	scanf("%d%d", &n, &m);
	for(i = 0; i <= n; i++){
		scanf("%s", s);
		for(j = 0; j < 1<<i; j++){
			if(s[j] == '1') f[i][j] = 1;
		}
	}
	for(i = 1; i <= n; i++){
		for(j = 0; j < 1<<i; j++){
			int p = (j>>(i-1))&1;
			for(k = 0; k < i; k++) if(((j>>i-k-1)&1) != p) break;
			if(k == i) g[i][j] = -1;
			else g[i][j] = k;
		}
	}
	int ansi = 0, ansj;
	for(i = 0; i <= n; i++){
		for(j = 1; i+j <= n; j++){
			for(k = 0; k < 1<<i+j; k++){
				f[i][k>>j] += f[i+j][k];
				int p = g[j][k&((1<<j)-1)];
				if(p != -1) f[i+j-p][(k>>j<<(j-p))|(k&((1<<(j-p))-1))] += f[i+j][k];
			}
		}
		for(j = (1<<i)-1; j >= 0; j--){
			if(f[i][j] >= m) ansi = i, ansj = j;
		}
	}
	for(i = ansi-1; i >= 0; i--) printf("%d", (ansj>>i)&1);
	printf("\n");
	return 0;
}