Problem:
Score : 2300 points

Problem Statement

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.

Sample Input 1

3 4
1
01
1011
01001110

Sample Output 1

10

The following strings belong to S: the empty string, 1, 00, 10, 11, 001, 100, 101 and 110.
The lexicographically smallest string among the longest strings that are subsequences of four or more of them is 10.

Sample Input 2

4 6
1
01
1011
10111010
1101110011111101

Sample Output 2

100

Sample Input 3

2 5
0
11
1111

Sample Output 3

The answer is the empty string.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define db double
#define pint pair<int,int>
#define mk(x,y) make_pair(x,y)
#define fir first
#define sec second
#define Rep(x,y,z) for(int x=y;x<=z;++x)
#define Red(x,y,z) for(int x=y;x>=z;--x)
using namespace std;
char buf[1<<12],*p1=buf,*p2=buf,nc;int ny;
//inline char gc() {return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<12,stdin),p1==p2)?EOF:*p1++;}
inline char gc(){return getchar();}
inline int read(){
	int x=0;ny=1;while(nc=gc(),(nc<48||nc>57)&&nc!=EOF)if(nc==45)ny=-1;if(nc<0)return nc;
	x=nc-48;while(nc=gc(),47<nc&&nc<58&&nc!=EOF)x=(x<<3)+(x<<1)+(nc^48);return x*ny;
}int n,k,nxt[2][1<<21],F[21][1<<21],Lg[1<<21],tot;
char s[1<<20];
int main(){
//	freopen("std.in","r",stdin);
//	freopen("std.out","w",stdout);
	n=read(),k=read();
	Rep(i,0,n){scanf("%s",s);for(int j=0;j<1<<i;j++)if(s[j]==49)F[i][1<<i^j]=1;}
	Lg[0]=-1;for(int i=1;i<2<<n;i++)Lg[i]=Lg[i>>1]+1;
	for(int s=2;s<2<<n;s++){
		if(s>>(Lg[s]-1)&1)nxt[1][s]=s^1<<Lg[s],nxt[0][s]=nxt[0][nxt[1][s]];
		else nxt[0][s]=s^3<<Lg[s]-1,nxt[1][s]=nxt[1][nxt[0][s]];
	}Red(i,n,1)for(int s=1<<i;s<2<<n;s++)if(F[i][s]){
		int A=s>>i,B=(s&((1<<i)-1))^1<<i;
		if(nxt[0][B])F[Lg[nxt[0][B]]][(A<<1^1)<<Lg[nxt[0][B]]^nxt[0][B]]+=F[i][s];
		if(nxt[1][B])F[Lg[nxt[1][B]]][(A<<1^0)<<Lg[nxt[1][B]]^nxt[1][B]]+=F[i][s];
		F[0][A]+=F[i][s];
	}int Ans=1;
	for(int i=2;i<2<<n;i++)if(F[0][i]>=k&&Lg[i]>Lg[Ans])Ans=i;
	int j=n;for(;~Ans>>j&1;j--);for(;j--;)cout<<(Ans>>j&1);
	return 0;
}