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 a[21][(1 << 20) + 1], f[21][(1 << 20) + 1];

int main() {
	ios::sync_with_stdio(false);
	
	cin >> n >> m;
	for (int i = 0; i <= n; ++i) {
		string s;
		cin >> s;
		for (int j = 0; j < (1 << i); ++j) {
			a[i][j] = s[j] - '0';
		}
	}
	
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j < (1 << i); ++j) {
			int b = j >> (i - 1) & 1, r = 0;
			while (r < i && (j >> (i - r - 1) & 1) == b) {
				++r;
			}
			f[i][j] = r == i ? -1 : r;
		}
	}
	
	int len = 0, ans = 0;
	for (int i = 0; i <= n; ++i) {
		for (int j = 1; j <= n - i; ++j) {
			for (int k = 0; k < (1 << (i + j)); ++k) {
				a[i][k >> j] += a[i + j][k];
				int tmp = f[j][k & ((1 << j) - 1)];
				if (!~tmp) continue;
				a[i + j - tmp][((k >> j) << (j - tmp)) | (k & ((1 << (j - tmp)) - 1))] += a[i + j][k];
			}
		}
		for (int j = (1 << i) - 1; j >= 0; --j) {
			if (a[i][j] >= m) {
				len = i;
				ans = j;
			}
		}
	}
	
	for (int i = len - 1; i >= 0; --i) {
		printf("%d", ans >> i & 1);
	}
}