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>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 998244353;

int n, k, cnt;
char hv[25][1048576];
int cdp[1048576][25], ndp[1048576][25];

int main()
{
	scanf("%d%d", &n, &cnt);
	for(int i = 0; i <= n; i ++) scanf("%s", hv[i]);
	
	rep(i, 1 << n) if(hv[n][i] == '1') for(int j = 0; j <= n; j ++) cdp[i][j] = 1;
	rep(i, 1 << n) if(cdp[i][0] >= cnt) {
		for(int j = n - 1; j >= 0; j --) printf("%d", i >> j & 1);
		printf("\n");
		return 0;
	}
	for(int i = n - 1; i >= 0; i --) {
		rep(j, 1 << i) for(int k = 0; k <= i; k ++) ndp[j][k] = 0;
		rep(j, 1 << i) for(int k = i; k >= 0; k --)
		if(k == i) ndp[j][k] = cdp[j][k + 1] + cdp[j | 1 << i][k + 1];
		else {
			int cj = j & (1 << k) - 1;
			cj |= (j ^ cj) << 1;
			cj |= ~j & 1 << k;
			ndp[j][k] = cdp[cj][k + 1] + ndp[j][k + 1];
		}
		rep(j, 1 << i) if(hv[i][j] == '1') for(int k = 0; k <= i; k ++) ndp[j][k] ++;
		
		rep(j, 1 << i) for(int k = 0; k <= i; k ++) cdp[j][k] = ndp[j][k];
		
		rep(j, 1 << i) if(cdp[j][0] >= cnt) {
			for(int k = i - 1; k >= 0; k --) printf("%d", j >> k & 1);
			printf("\n");
			return 0;
		}
	}
	
	return 0;
}