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 ctz __builtin_ctz

const int N = 27, N2 = 2097154;

int n, K;
int f[N2];
char s[N2];

inline int next0(int v) {return v & (v + 1) ? ctz(~v) : -1;}
inline int next1(int v) {return v & (v - 1) ? ctz(v) : -1;}
int (*const next[2])(int) = {next0, next1};

int main() {
	int i, j, S, ALL, lo, hi;
	scanf("%d%d", &n, &K), ALL = ~(-2 << n);
	for (i = 0; i <= n; ++i)
		for (scanf("%s", s + (j = 1 << i)); j < 2 << i; ++j) f[j] = s[j] & 1;
	for (i = 0; i < n; ++i)
		for (S = 2 << i; S <= ALL; ++S) {
			lo = S & ~(-1 << i), hi = S >> i, f[1 << i | lo] += f[S];
			if (~(j = next[~hi & 1](hi))) f[hi >> j << i | lo] += f[S];
		}
	for (i = n; i; --i)
		for (j = 0; j < 1 << i; ++j)
			if (f[1 << i | j] >= K) {for (; i; putchar((j >> --i & 1) | 48)); return putchar(10), 0;}
	return 0;
}