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<cstdio>
#include<algorithm>
using namespace std;
int n, K, M[1<<22], rb, rnum = -1;
int D[21][1 << 21];
char p[(1 << 20) + 2];
int Num(int dep, int a, int num, int b) {
	return ((a + 1) << (n - dep + 1)) - (1 << (num + 1)) + b;
}
int main() {
	int i, j;
	scanf("%d%d", &n, &K);
	for (i = 1; i <= (1 << (n+1)); i++) {
		M[i] = M[i - 1];
		while ((1 << M[i]) < i)M[i]++;
	}
	for (i = 0; i <= n; i++) {
		scanf("%s", p);
		for (j = 0; j < (1 << i); j++) {
			if (p[j] == '1') {
				D[i][Num(i, j, 0, 0)]++;
			}
		}
	}

	for (i = n - 1; i >= 0; i--) {
		for (j = 0; j < (1 << (n + 1)); j++) {
			if ((j + 1) % (1 << (n + 1 - i)) == 0)continue;
			int a = j >> (n - i + 1);
			int t = ((a + 1) << (n - i + 1)) - j;
			int num = M[t] - 1;
			if (num < 0) {
				printf("%d %d\n", i, j);
			}
			int b = (1 << (num + 1)) - t;
			int ck = -1;
			if (num) ck = b >> (num - 1);
			for (int k = 0; k < 2; k++) {
				if (ck == k) {
					D[i][j] += D[i+1][Num(i + 1, a * 2 + k, num - 1, b & ((1 << (num - 1)) - 1)  )];
				}
				else if(i+1+num<=n){
					D[i][j] += D[i + 1][Num(i + 1, a * 2 + k, num, b)];
				}
			}
		}
	}
	i = 0;
	for (j = 0; j < (1 << (n + 1)) - 1; j++) {
		if (D[0][j] >= K) {
			int a = j >> (n - i + 1);
			int t = ((a + 1) << (n - i + 1)) - j;
			int num = M[t] - 1;
			int b = (1 << (num + 1)) - t;
			if (rnum < num)rnum = num, rb = b;
		}
	}
	for (i = rnum - 1; i >= 0; i--) {
		printf("%d", (rb >> i) & 1);
	}

	return 0;
}