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>

const int MN = 21;

int N, K, s[1 << MN], tr[1 << MN][2];
int f[MN][1 << MN], Ans;

int main() {
	scanf("%d%d", &N, &K);
	for (int i = 0; i <= N; ++i)
		for (int x = 0; x < 1 << i; ++x)
			scanf("%1d", &f[i][1 << i ^ x]);
	s[0] = -1;
	for (int A = 1; A < 2 << N; ++A) s[A] = s[A >> 1] + 1;
	for (int A = 2; A < 2 << N; ++A)
		if (A >> (s[A] - 1) & 1) {
			tr[A][1] = A ^ 1 << s[A];
			tr[A][0] = tr[tr[A][1]][0];
		} else {
			tr[A][0] = A ^ 3 << (s[A] - 1);
			tr[A][1] = tr[tr[A][0]][1];
		}
	for (int i = N; i >= 1; --i) {
		for (int A = 1 << i; A < 2 << N; ++A) {
			if (!f[i][A]) continue;
			int _ = A >> i, B = (A & ((1 << i) - 1)) ^ 1 << i;
			if (tr[B][0]) f[s[tr[B][0]]][(_ << 1 ^ 1) << s[tr[B][0]] ^ tr[B][0]] += f[i][A];
			if (tr[B][1]) f[s[tr[B][1]]][(_ << 1 ^ 0) << s[tr[B][1]] ^ tr[B][1]] += f[i][A];
			f[0][_] += f[i][A];
		}
	}
	Ans = 1;
	for (int A = 2; A < 2 << N; ++A) if (f[0][A] >= K && s[A] > s[Ans]) Ans = A;
	int j = N;
	while (~Ans >> j & 1) --j;
	while (j--) printf("%d", Ans >> j & 1);
	puts("");
	return 0;
}