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 rep(i,a,b) for (int i = (a); i <= (b); ++i)
#define drep(i,a,b) for (int i = (a); i >= (b); --i)
#define grep(i,u) for (int i = head[u],v = e[i].v; i; v = e[i = e[i].nxt].v)
#define il inline
#define LL long long
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define pll pair<LL,LL>
#define fi first
#define se second
using namespace std;
il int read() {
	int res = 0,f = 1; char ch = getchar(); while (!isdigit(ch)) {if (ch == '-') f = -f; ch = getchar();}
	while (isdigit(ch)) {res = res*10+ch-'0'; ch = getchar();} return res*f;
}
namespace qiqi {
	const int N = 21; int n,m,f[N][1<<N],tr[1<<N][2],s[1<<N];
	il void print(int x) {drep(i,s[x],0) putchar('0'+((x>>i)&1)); putchar(' ');}
	void main() {
		int x,k; n = read(); m = read(); char ch = getchar(); rep(i,0,n) {while (ch != '0' && ch != '1') ch = getchar(); rep(j,0,(1<<i)-1) {f[i][j|(1<<i)] = ch-'0'; ch = getchar();}}
		s[0] = -1; rep(i,1,(2<<n)-1) if (s[i] = s[i>>1]+1) {if ((i>>(s[i]-1))&1) tr[i][0] = tr[tr[i][1] = i^(1<<s[i])][0]; else tr[i][1] = tr[tr[i][0] = i^(3<<(s[i]-1))][1];}
		drep(i,n,1) rep(j,1<<i,(2<<n)-1) if (f[i][j]) {k = (j&((1<<i)-1))|(1<<i); f[0][x=j>>i] += f[i][j]; if (tr[k][0]) f[s[tr[k][0]]][(((x<<1)|1)<<s[tr[k][0]])^tr[k][0]] += f[i][j]; if (tr[k][1]) f[s[tr[k][1]]][(x<<(s[tr[k][1]]+1))|tr[k][1]] += f[i][j];}
		x = 1; rep(i,2,(2<<n)-1) if (f[0][i] >= m && s[i] > s[x]) x = i; drep(i,s[x]-1,0) putchar('0'+((x>>i)&1));
	}
}
int main() {qiqi::main(); return 0;}