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;
vector<int> f[maxn][maxn];
int main(){
	for(int i = 0; i < maxn; i++){
		for(int j = 0; j <= i; j++){
			f[i][j].resize(1<<i);
		}
	}
	int n, k; cin >> n >> k;
	for(int i = 0; i <= n; i++){
		string s; cin >> s;
		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 z = 0; z < (1<<i); z++){
				int val = f[i][j][z];
				if(val == 0) continue;
				{
					f[j][j][z&((1<<j)-1)] += val;
				}
				{
					for(int rot = 0; rot < 2; rot++){
						int pos = -1;
						for(int t = j; t < i; t++){
							if(((z>>t)&1) == rot){
								pos = t;
								break;
							}
						}
						if(pos != -1){
							int nz = (z&((1<<j)-1)) + ((z>>pos)<<j);
							f[i - pos + j][j + 1][nz] += val;
						}
					}
				}
			}
		}
		for(int j = 0; j < (1<<i); j++){
			if(f[i][i][j] >= k){
				for(int z = i - 1; z >= 0; z--) cout << ((j>>z)&1);
				cout << endl;
				return 0;
			}
		}
	}
	return 0;
}