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;
int n, m;
int f[21][1048577];
int g[21][1048577];
char s[1048577];
int zi, zj;
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++) {
			f[i][j] = s[j] - '0';
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < 1 << i; j++) {
			int p = i - 1, b = (j >> (i - 1)) & 1;
			while (p >= 0 &&  b == ((j >> p) & 1)) {
				p--;
			}
			g[i][j] = p;
		}
	}
	for (int i = 0; i <= n; i++) {
		for (int j = 1; i + j <= n; j++) {
			for (int 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 >= 0) {
					f[i + p + 1][(k & ((1 << (p + 1)) - 1)) | (k >> j << (p + 1))] += f[i + j][k];
				}
			}
		}
		for (int j = (1 << i) - 1; j >= 0; j--) {
			if (f[i][j] >= m) {
				zi = i;
				zj = j;
			}
		}
	}
	for (int i = zi - 1; i >= 0; i--) {
		printf("%d", zj >> i & 1);
	}
	printf("\n");
	return 0;
}