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;

const int maxn = 21, maxm = 1 << maxn;
int n, k, dp[maxn + 3][maxm + 3];
char s[maxm + 3];

int main() {
	scanf("%d %d", &n, &k);
	for (int i = 0; i <= n; i++) {
		scanf("%s", s);
		for (int j = 0; j < 1 << i; j++) {
			dp[i][1 << i | j] = s[j] ^ '0';
		}
	}
	for (int i = n; i; i--) {
		for (int j = i; j <= n; j++) {
			int b = 1 << j;
			for (int k = 0; k < b; k++) {
				int x = dp[i][k | b], p = -1, q = -1;
				for (int t = i; t; t--) {
					if (~k >> (t - 1) & 1) {
						p = t;
						break;
					}
				}
				for (int t = i; t; t--) {
					if (k >> (t - 1) & 1) {
						q = t;
						break;
					}
				}
				dp[0][(k | b) >> i] += x;
				if (p != -1) {
					dp[p - 1][(k & ((1 << (p - 1)) - 1)) | (((k >> i) << 1 | 0) << (p - 1)) | (b >> (i - p))] += x;
				}
				if (q != -1) {
					dp[q - 1][(k & ((1 << (q - 1)) - 1)) | (((k >> i) << 1 | 1) << (q - 1)) | (b >> (i - q))] += x;
				}
			}
		}
	}
	for (int i = n; i >= 0; i--) {
		for (int j = 1 << i; j < 1 << (i + 1); j++) {
			if (dp[0][j] >= k) {
				for (int t = i - 1; t >= 0; t--) {
					printf("%d", j >> t & 1);
				}
				puts("");
				return 0;
			}
		}
	}
	return 0;
}