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 = 25;
const int MAXS = 1 << 21;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
char s[MAXS];
int n, k, dp[MAXN][MAXS];
int main() {
	read(n), read(k);
	for (int i = 0, bit = 1 << i; i <= n; i++, bit <<= 1) {
		scanf("\n%s", s);
		for (int j = 0; j <= bit - 1; j++)
			dp[i][j + bit] = s[j] == '1';
	}
	for (int i = n, bit = 1 << i; i >= 1; i--, bit >>= 1) {
		for (int j = i, bjt = 1 << j; j <= n; j++, bjt <<= 1)
		for (int k = 0; k <= bjt - 1; k++) {
			int tmp = dp[i][k + bjt], p = -1, q = -1;
			for (int t = i; t >= 1; t--)
				if (k & (1 << (t - 1))) {
					p = t;
					break;
				}
			for (int t = i; t >= 1; t--)
				if ((k & (1 << (t - 1))) == 0) {
					q = t;
					break;
				}
			dp[0][(k + bjt) >> i] += tmp;
			int s = k & ((1 << i) - 1), t = k >> i;
			if (p != -1) dp[p - 1][(s & ((1 << (p - 1)) - 1)) + ((2 * t + 1) << (p - 1)) + (bjt >> (i - p))] += tmp;
			if (q != -1) dp[q - 1][(s & ((1 << (q - 1)) - 1)) + ((2 * t) << (q - 1)) + (bjt >> (i - q))] += tmp;
		}
	}
	for (int i = n, bit = 1 << i; i >= 0; i--, bit >>= 1) {
		for (int j = 0; j <= bit - 1; j++)
			//printf("%d %d %d\n", i, j, dp[0][j + bit]);
			if (dp[0][j + bit] >= k) {
				for (int k = i - 1; k >= 0; k--)
					if (j & (1 << k)) putchar('1');
					else putchar('0');
				puts("");
				return 0;
			}
	}
	assert(false);
	return 0;
}