#include<bits/stdc++.h>
using namespace std;
#define maxn 25
#define maxs 2100000
template <typename T>inline T read()
{
    register T sum=0;
    register char cc=getchar();
    int sym=1;
    while(cc!='-'&&(cc>'9'||cc<'0'))cc=getchar();
    if(cc=='-')sym=-1,cc=getchar();
    sum=sum*10+cc-'0';
    cc=getchar();
    while(cc>='0'&&cc<='9')sum=sum*10+cc-'0',cc=getchar();
    return sym*sum;
}
template <typename T>inline T read(T &a)
{
    a=read<T>();
    return a;
}
template <typename T,typename... Others> inline void read(T& a, Others&... b)
{
    a=read(a);
	read(b...);
}
int n,m,f[maxs][maxn];
char str[maxs];
int main()
{
	read(n,m);
	for(int i=0;i<=n;i++)
	{
		scanf("%s",str);
		for(int j=0;j<(1<<i);j++)
			f[(1<<i)|j][i]=str[j]=='1';
	}
	for(int i=n;i>=0;i--)
		for(int s=0;s<(1<<i);s++)
			for(int j=i;j>=0;j--)
				if(f[(1<<i)|s][j])
				{
					int v=f[(1<<i)|s][j];
					int s1=s>>j,s2=s&((1<<j)-1);
					if(j)f[(1<<(i-j))|s1][0]+=v;
					if(s2)
					{
						int p=j-1;
						while(((s2>>p)&1)==0)
							p-=1;
						f[(1<<(i-(j-1-p)))|s1<<(p+1)|(1<<p)|(s2&((1<<p)-1))][p]+=v;
					}
					if(s2!=(1<<j)-1)
					{
						int p=j-1;
						while(((s2>>p)&1)==1)
							p-=1;
						f[(1<<(i-(j-1-p)))|s1<<(p+1)|(s2&((1<<p)-1))][p]+=v;
					}
				}
	for(int i=n;i>=0;i--)
		for(int j=0;j<(1<<i);j++)
			if(f[(1<<i)|j][0]>=m)
			{
				for(int k=i-1;k>=0;k--)
					putchar(((j>>k)&1)+'0');
				return 0;
			}
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define maxn 25
#define maxs 2100000
template <typename T>inline T read()
{
    register T sum=0;
    register char cc=getchar();
    int sym=1;
    while(cc!='-'&&(cc>'9'||cc<'0'))cc=getchar();
    if(cc=='-')sym=-1,cc=getchar();
    sum=sum*10+cc-'0';
    cc=getchar();
    while(cc>='0'&&cc<='9')sum=sum*10+cc-'0',cc=getchar();
    return sym*sum;
}
template <typename T>inline T read(T &a)
{
    a=read<T>();
    return a;
}
template <typename T,typename... Others> inline void read(T& a, Others&... b)
{
    a=read(a);
	read(b...);
}
int n,m,f[maxs][maxn];
char str[maxs];
int main()
{
	read(n,m);
	for(int i=0;i<=n;i++)
	{
		scanf("%s",str);
		for(int j=0;j<(1<<i);j++)
			f[(1<<i)|j][i]=str[j]=='1';
	}
	for(int i=n;i>=0;i--)
		for(int s=0;s<(1<<i);s++)
			for(int j=i;j>=0;j--)
				if(f[(1<<i)|s][j])
				{
					int v=f[(1<<i)|s][j];
					int s1=s>>j,s2=s&((1<<j)-1);
					if(j)f[(1<<(i-j))|s1][0]+=v;
					if(s2)
					{
						int p=j-1;
						while(((s2>>p)&1)==0)
							p-=1;
						f[(1<<(i-(j-1-p)))|s1<<(p+1)|(1<<p)|(s2&((1<<p)-1))][p]+=v;
					}
					if(s2!=(1<<j)-1)
					{
						int p=j-1;
						while(((s2>>p)&1)==1)
							p-=1;
						f[(1<<(i-(j-1-p)))|s1<<(p+1)|(s2&((1<<p)-1))][p]+=v;
					}
				}
	for(int i=n;i>=0;i--)
		for(int j=0;j<(1<<i);j++)
			if(f[(1<<i)|j][0]>=m)
			{
				for(int k=i-1;k>=0;k--)
					putchar(((j>>k)&1)+'0');
				return 0;
			}
    return 0;
}

```