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, dp[42000000];
vector<int> id[21];
char s[21][1145141];
int main() {
	scanf("%d%d", &n, &k);
	for (int i = 0; i <= n; i++) {
		scanf(" %s", s[i]);
	}
	int tot = 0;
	for (int i = 0; i <= n; i++) {
		id[i].resize((i + 1) * (1 << i));
		for (int j = 0; j < (i + 1) * (1 << i); j++) {
			id[i][j] = tot++;
		}
	}
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j < (1 << i); j++) {
			if (s[i][j] == '1')
				dp[id[i][j * (i + 1)]] = 1;
		}
	}
	for (int i = n; i >= 0; i--) {
		for (int j = 0; j < (1 << i); j++) {
			for (int p = 0; p < i; p++) if (dp[id[i][j * (i + 1) + p]]) {
				dp[id[p][(j >> (i - p)) * (p + 1) + p]] += dp[id[i][j * (i + 1) + p]];
				int fnd = -1;
				for (int q = p + 1; q <= i; q++) {
					if ((j >> (i - q)) & 1) {
						fnd = q;
						break;
					}
				}
				if (fnd != -1) {
					dp[id[p + i - fnd + 1][(((j >> (i - p)) << (i - fnd + 1)) | (j & ((1 << (i - fnd + 1)) - 1))) * (p + i - fnd + 2) + p + 1]] += dp[id[i][j * (i + 1) + p]];
				}
				fnd = -1;
				for (int q = p + 1; q <= i; q++) {
					if (!((j >> (i - q)) & 1)) {
						fnd = q;
						break;
					}
				}
				if (fnd != -1) {
					dp[id[p + i - fnd + 1][(((j >> (i - p)) << (i - fnd + 1)) | (j & ((1 << (i - fnd + 1)) - 1))) * (p + i - fnd + 2) + p + 1]] += dp[id[i][j * (i + 1) + p]];
				}
			}
			if (dp[id[i][j * (i + 1) + i]] >= k) {
				for (int p = i - 1; p >= 0; p--) {
					putchar('0' + ((j >> p) & 1));
				}
				putchar('\n');
				return 0;
			}
		}
	}
	return 0;
}