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 <iostream>

using namespace std;

int dp[21][(1 << 21)], f[21][(1 << 21)];

int main ()
{
int n, k;
cin >> n >> k;
for(int i = 0 ; i <= n ; ++i){
	string s;
	cin >> s;
	for(int j = 0 ; j < s.size() ; ++j){
		if(s[j] == '1'){
			dp[i][j] = 1;
		}
	}
}
for(int i = 1 ; i <= n ; ++i){
	for(int j = 0 ; j < (1 << i) ; ++j){
		int u = (j >> (i - 1)) & 1;
		int br = 0;
		while(br < i && ((j >> (i - br - 1)) & 1) == u){
			br++;
		}
		if(br == i){
			f[i][j] = -1;
		}
		else{
			f[i][j] = br;
		}
	}
}
for(int i = 0 ; i <= n ; ++i){
	for(int j = i + 1 ; j <= n ; ++j){
		int mask = (1 << (j - i)) - 1;
		for(int w = 0 ; w < (1 << j) ; ++w){
			dp[i][w >> (j - i)] += dp[j][w];
			int x = f[j - i][w & mask];
			if(x != -1){
				dp[j - x][((w >> (j - i)) << (j - i - x)) + (w & ((1 << (j - i - x)) - 1))] += dp[j][w];
			}
		}
	}
}
int sol, vel = -1;
for(int i = 0 ; i <= n ; ++i){
	for(int j = 0 ; j < (1 << i) ; ++j){
		if(dp[i][j] >= k){
			if(i > vel){
				vel = i;
				sol = j;
			}
		}
	}
}
for(int i = vel - 1 ; i >= 0 ; --i){
	cout << ((sol >> i) & 1);
}

return 0;
}