#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#define llint long long
#define inf 1e18
#define rep(x, s, t) for(llint (x) = (s); (x) < (t); (x)++)
#define Rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))

using namespace std;
typedef pair<llint, llint> P;

llint n, m;
string s;
bool b[1<<21];
int dp[1<<21][21];

string enc(llint mask, llint len)
{
	string ret;
	for(int i = 0; i < len; i++){
		ret += mask%2 + '0';
		mask /= 2;
	}
	reverse(ret.begin(), ret.end());
	return ret;
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> m;
	for(int i = 0; i <= n; i++){
		cin >> s;
		llint M = 1<<i;
		for(int j = 0; j < M; j++){
			if(s[j]%2) b[M+j] = true;
		}
	}
	
	for(int i = n; i >= 0; i--){
		llint M = 1<<i;
		for(int j = 0; j < M; j++){
			llint x = M+j;
			for(int k = i; k >= 0; k--){
				if(b[x]) dp[x][k] = 1;
				if(i == n) continue;
				dp[x][k] += dp[x*2][0] + dp[x*2+1][0];
				for(int l = 0; l < k; l++){
					llint y = (x >> (l+1)) << (l+2);
					y |= (((x >> l) & 1) ^ 1) << (l+1);
					y |= x & ((1<<(l+1))-1);
					dp[x][k] += dp[y][l+1];
				}
			}
		}
	}
	
	string ans;
	for(int i = n; i >= 0; i--){
		llint M = 1<<i;
		for(int j = 0; j < M; j++){
			if(dp[M+j][i] >= m){
				ans = enc(j, i);
				goto end;
			}
		}
	}
	end:;
	cout << ans << endl;
	
	return 0;
}
### Prompt

Please create a solution in CPP to the following 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

### Response

```cpp
#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#define llint long long
#define inf 1e18
#define rep(x, s, t) for(llint (x) = (s); (x) < (t); (x)++)
#define Rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))

using namespace std;
typedef pair<llint, llint> P;

llint n, m;
string s;
bool b[1<<21];
int dp[1<<21][21];

string enc(llint mask, llint len)
{
	string ret;
	for(int i = 0; i < len; i++){
		ret += mask%2 + '0';
		mask /= 2;
	}
	reverse(ret.begin(), ret.end());
	return ret;
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> m;
	for(int i = 0; i <= n; i++){
		cin >> s;
		llint M = 1<<i;
		for(int j = 0; j < M; j++){
			if(s[j]%2) b[M+j] = true;
		}
	}
	
	for(int i = n; i >= 0; i--){
		llint M = 1<<i;
		for(int j = 0; j < M; j++){
			llint x = M+j;
			for(int k = i; k >= 0; k--){
				if(b[x]) dp[x][k] = 1;
				if(i == n) continue;
				dp[x][k] += dp[x*2][0] + dp[x*2+1][0];
				for(int l = 0; l < k; l++){
					llint y = (x >> (l+1)) << (l+2);
					y |= (((x >> l) & 1) ^ 1) << (l+1);
					y |= x & ((1<<(l+1))-1);
					dp[x][k] += dp[y][l+1];
				}
			}
		}
	}
	
	string ans;
	for(int i = n; i >= 0; i--){
		llint M = 1<<i;
		for(int j = 0; j < M; j++){
			if(dp[M+j][i] >= m){
				ans = enc(j, i);
				goto end;
			}
		}
	}
	end:;
	cout << ans << endl;
	
	return 0;
}
```