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 fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0){putchar('-');x = -x;}
	if(x >= 10) out(x / 10);
	putchar('0' + x % 10);
}

int N,K,cnt;
char s[25][(1 << 20) + 5];
vector<int> v[25];
int dp[42000005];
int p0[25],p1[25];
void Solve() {
	read(N);read(K);
	for(int i = 0 ; i <= N ; ++i) scanf("%s",s[i]);
	for(int i = 1 ; i <= N ; ++i) {
		v[i].resize((i + 1) * (1 << i));
		int s = v[i].size();
		for(int j = 0 ; j < s ; ++j) v[i][j] = ++cnt;
	}
	for(int i = 1 ; i <= N ; ++i) {
		for(int j = 0 ; j < (1 << i) ; ++j) {
			if(s[i][j] == '1') {
				dp[v[i][j * (i + 1) + i]]++;
			}
		}
	}
	for(int i = N ; i >= 1 ; --i) {
		for(int S = 0 ; S < (1 << i) ; ++S) {
			p0[0] = -1,p1[0] = -1;
			for(int j = 0 ; j <= i ; ++j) {
				p1[j + 1] = p1[j];p0[j + 1] = p0[j];
				if(S >> j & 1) p1[j + 1] = j;
				else p0[j + 1] = j;
			}
			for(int j = i ; j >= 0 ; --j) {
				int u = v[i][S * (i + 1) + j];
				int b = S & ((1 << j) - 1),f = S >> j;
				if(dp[u]) {
					if(p0[j] != -1) {
						int p = ((f << 1) << p0[j]) + (S & (1 << p0[j]) - 1);
						int l = p0[j] + 1 + i - j;
						dp[v[l][p * (l + 1) + p0[j]]] += dp[u];
					}
					if(p1[j] != -1) {
						int p = ((f << 1 | 1) << p1[j]) + (S & (1 << p1[j]) - 1);
						int l = p1[j] + 1 + i - j;
						dp[v[l][p * (l + 1) + p1[j]]] += dp[u];
					}
					if(j != 0 && j != i) {
						int l = i - j;
						dp[v[l][f * (l + 1)]] += dp[u];
					}
				}
			}
		}
	}
	for(int i = N ; i >= 1 ; --i) {
		for(int j = 0 ; j < (1 << i) ; ++j) {
			if(dp[v[i][j * (i + 1)]] >= K) {
				for(int k = i - 1; k >= 0 ; --k) {
					putchar('0' + ((j >> k) & 1));
				}
				enter;
				return;
			}
		}
	}
	puts("");
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	Solve();
}