/*
* @Author: wxyww
* @Date: 2020-03-19 08:57:18
* @Last Modified time: 2020-03-19 10:05:44
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1 << 21;
ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
int *f[N],trans[N][2];
char s[N];
int main() {
	int n = read(),K = read();
	f[1] = new int[1 << (n + 1)];
	f[1][0] = 0;
	for(int i = 0;i <= n;++i) {
		int p = 1 << i;
		scanf("%s",s);
		for(int j = 0;j < p;++j) f[1][p + j] = s[j] - '0';
		for(int j = 0;j < p;++j) trans[p + j][1] = j;
		trans[p + p - 1][0] = 0;
		for(int j = 1;j < p;++j) trans[p + p - j - 1][0] = j ^ ((1 << (31 - __builtin_clz(j))) - 1);
	}
	
	// puts("!!!");

	int ans = 1;
	for(int i = 1;i <= n;++i) {
		int p = 1 << i;
		for(int j = p + p - 1;j >= p;--j) {
			f[j] = new int[1 << (n - i + 1)];
			memset(f[j],0,sizeof(int) << (n - i + 1));
			for(int k = 1;k < (1 << (n - i + 2));++k) f[j][trans[k][j & 1]] += f[j >> 1][k];
			f[j][0] = 0;
			if(accumulate(f[j],f[j] + (1 << (n - i + 1)),0) >= K) ans = j;
		}
	}
	// puts("!!!");
	for(int i = 30 - __builtin_clz(ans);i >= 0;--i) printf("%d",ans >> i & 1);

	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
/*
* @Author: wxyww
* @Date: 2020-03-19 08:57:18
* @Last Modified time: 2020-03-19 10:05:44
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<bitset>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1 << 21;
ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
int *f[N],trans[N][2];
char s[N];
int main() {
	int n = read(),K = read();
	f[1] = new int[1 << (n + 1)];
	f[1][0] = 0;
	for(int i = 0;i <= n;++i) {
		int p = 1 << i;
		scanf("%s",s);
		for(int j = 0;j < p;++j) f[1][p + j] = s[j] - '0';
		for(int j = 0;j < p;++j) trans[p + j][1] = j;
		trans[p + p - 1][0] = 0;
		for(int j = 1;j < p;++j) trans[p + p - j - 1][0] = j ^ ((1 << (31 - __builtin_clz(j))) - 1);
	}
	
	// puts("!!!");

	int ans = 1;
	for(int i = 1;i <= n;++i) {
		int p = 1 << i;
		for(int j = p + p - 1;j >= p;--j) {
			f[j] = new int[1 << (n - i + 1)];
			memset(f[j],0,sizeof(int) << (n - i + 1));
			for(int k = 1;k < (1 << (n - i + 2));++k) f[j][trans[k][j & 1]] += f[j >> 1][k];
			f[j][0] = 0;
			if(accumulate(f[j],f[j] + (1 << (n - i + 1)),0) >= K) ans = j;
		}
	}
	// puts("!!!");
	for(int i = 30 - __builtin_clz(ans);i >= 0;--i) printf("%d",ans >> i & 1);

	return 0;
}
```