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;
typedef long long ll;
typedef double db;
const int N=22,M=1000010,P=1e9+7;
const int inf=0x3f3f3f3f;
const int INF=0xcfcfcfcf;
const double eps=1e-9,pi=asin(1)*2;
inline ll read();
#define cmax(a,b) a=max(a,b)
#define cmin(a,b) a=min(a,b)
inline int ADD(int a,int b) { return a+b>=P?a+b-P:a+b; }
inline int MINUS(int a,int b) { return a-b<0?a-b+P:a-b; }
#define plus(a,b) a=ADD(a,b)
#define minus(a,b) a=MINUS(a,b)
#define mul(a,b) a=(1ll*a*(b))%P
#define mem(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pll pair<ll,ll>
#define mp(x,y) make_pair(x,y)

int n,K;
int dp[(1<<N)|10][N];
char s[(1<<N)|10];

#ifdef FILE
const int RS=1<<20;
char buf[RS],*p1=buf,*p2=buf;
#define getchar() p1==p2&&(p2=(p1=buf)+fread(buf,1,RS,stdin),p1==p2)?EOF:*p1++
#endif
inline ll read()
{
	ll s=0;
	bool flag=false;
	char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') flag=true;
	for(;'0'<=ch&&ch<='9';ch=getchar()) s=(s<<3)+(s<<1)+(ch^'0');
	if(flag) return -s;
	return s;
}

int main()
{
#ifdef FILE
	freopen(FILE ".in","r",stdin);
	freopen(FILE ".out","w",stdout);
#endif
	n=read(),K=read();
	for(int i=0;i<=n;i++)
	{
		scanf("%s",s);
		for(int j=0;j<(1<<i);j++) dp[(1<<i)|j][i]=(s[j]=='1');
	}
	for(int i=n;i>=0;i--)
	{
		for(int j=0;j<(1<<i);j++)
		{
			for(int k=i;k>=0;k--)
			{
				int res=dp[(1<<i)|j][k],S=j>>k,T=j&((1<<k)-1);
				if(!res) continue;
				if(k) dp[(1<<i-k)|S][0]+=res;
				// Æ¥Åä³É¹¦, Ìõ¼þÊÇ k>0, ·ñÔò»áÓÐ (S,\empty)->(S,\empty)
				if(T!=((1<<k)-1)) // Æ¥Åä 0, Ìõ¼þÊÇ T ÖÐ±ØÐë´æÔÚ 0 
				{
					int p=k-1;
					while(T&(1<<p)) p--;
					dp[(1<<i-k+1+p)|(S<<p+1)|(T&(1<<p)-1)][p]+=res;
				}
				if(T) // Æ¥Åä 1, Ìõ¼þÊÇ T ÖÐ±ØÐë´æÔÚ 1 
				{
					int p=k-1;
					while(!(T&(1<<p))) p--;
					dp[(1<<i-k+1+p)|(S+S+1<<p)|(T&(1<<p)-1)][p]+=res;
				} 
			}
		}
	}
	for(int i=n;i>=0;i--)
	{
		for(int j=0;j<(1<<i);j++)
		{
/*
			for(int k=n;k>=0;k--)
				cerr<<"("<<i<<","<<j<<","<<k<<"): "<<dp[(1<<i)|j][k]<<"\n";
//*/
			if(dp[(1<<i)|j][0]>=K)
			{
				for(int l=i-1;l>=0;l--) putchar(((j>>l)&1)|'0');
				putchar('\n');
				return 0;
			}
		}
	}
	return 0;
}