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, k;
vector<int> f[21][21];
char s[1 << 20 | 1];

int main() {
	scanf("%d%d", &n, &k);
	for (int i = 0; i <= n; ++i) {
		scanf("%s", s);
		for (int j = 0; j <= i; ++j)
			f[i][j].resize(1 << i);
		for (int j = 0; j < 1 << i; ++j)
			if (s[j] == '1')
				f[i][0][j] = 1;
	}
	for (int i = n; i >= 0; --i) {
		for (int j = 0; j < i; ++j)
			for (int s = 0; s < 1 << i; ++s)
				if (f[i][j][s]) {
					int a = s & (1 << j) - 1, b = s >> j;
					f[j][j][a] += f[i][j][s];
					if (b != (1 << i - j) - 1) {
						int cnt = __builtin_ctz(~b);
						f[i - cnt][j + 1][b >> cnt << j | a] += f[i][j][s];
					}
					if (b) {
						int cnt = __builtin_ctz(b);
						f[i - cnt][j + 1][b >> cnt << j | a] += f[i][j][s];
					}
				}
		for (int s = 0; s < 1 << i; ++s)
			if (f[i][i][s] >= k) {
				for (int j = i - 1; j >= 0; --j)
					putchar("01"[s >> j & 1]);
				puts("");
				return 0;
			}
	}
}