#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i = (a); i < (b); ++i)
#define rrep(i,a,b) for(int i = (b); i --> (a);)
#define all(v) (v).begin(),(v).end()
#define trav(x,v) for(auto &x : v)
#define sz(v) int(v.size())
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);
	
	int n, k;
	cin >> n >> k;

	vector<string> xs(n+1);
	trav(x, xs) cin >> x;

	vector<vector<vi>> cnt(2, vector<vi>(n+1)); 
	rep(t,0,2) rep(i,0,n+1){
		cnt[t][i].resize(1<<i);
		rep(j,0,1<<i) cnt[t][i][j] = xs[i][j]-'0';
	}
	rep(l,0,n){
		int t = l%2, u = 1-t;

		rep(i,l,n+1) cnt[u][i].assign(1<<i,0);

		rep(m,l,n+1) rep(j,0,1<<m){
			cnt[u][l][j>>(m-l)] += cnt[t][m][j];	
		}
		
		rep(m,l+1,n+1) rep(j,0,1<<m){
			int jj = j & ((1<<(m-l))-1);
			int jjc = (~j) & ((1<<(m-l))-1);
			int zz = 31 - __builtin_clz(jjc),
			    oo = 31 - __builtin_clz(jj);

			if(jjc){
				cnt[u][l+1+zz][((j>>(m-l))<<(zz+1))|(j&((1<<zz)-1))]
					+= cnt[t][m][j];
				
			}
			if(jj){
				cnt[u][l+1+oo][((j>>(m-l))<<(oo+1))|(1<<oo)|(j&((1<<oo)-1))]
					+= cnt[t][m][j];
			}
		}

	}
	rrep(i,0,n+1){
		rep(j,0,1<<i) if(cnt[(i+1)%2][i][j] >= k){
			rrep(k,0,i) cout << ((j&(1<<k))>>k);
			cout << endl;
			return 0;
		}
	}
}
### Prompt

Create a solution in CPP for 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 <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i = (a); i < (b); ++i)
#define rrep(i,a,b) for(int i = (b); i --> (a);)
#define all(v) (v).begin(),(v).end()
#define trav(x,v) for(auto &x : v)
#define sz(v) int(v.size())
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);
	
	int n, k;
	cin >> n >> k;

	vector<string> xs(n+1);
	trav(x, xs) cin >> x;

	vector<vector<vi>> cnt(2, vector<vi>(n+1)); 
	rep(t,0,2) rep(i,0,n+1){
		cnt[t][i].resize(1<<i);
		rep(j,0,1<<i) cnt[t][i][j] = xs[i][j]-'0';
	}
	rep(l,0,n){
		int t = l%2, u = 1-t;

		rep(i,l,n+1) cnt[u][i].assign(1<<i,0);

		rep(m,l,n+1) rep(j,0,1<<m){
			cnt[u][l][j>>(m-l)] += cnt[t][m][j];	
		}
		
		rep(m,l+1,n+1) rep(j,0,1<<m){
			int jj = j & ((1<<(m-l))-1);
			int jjc = (~j) & ((1<<(m-l))-1);
			int zz = 31 - __builtin_clz(jjc),
			    oo = 31 - __builtin_clz(jj);

			if(jjc){
				cnt[u][l+1+zz][((j>>(m-l))<<(zz+1))|(j&((1<<zz)-1))]
					+= cnt[t][m][j];
				
			}
			if(jj){
				cnt[u][l+1+oo][((j>>(m-l))<<(oo+1))|(1<<oo)|(j&((1<<oo)-1))]
					+= cnt[t][m][j];
			}
		}

	}
	rrep(i,0,n+1){
		rep(j,0,1<<i) if(cnt[(i+1)%2][i][j] >= k){
			rrep(k,0,i) cout << ((j&(1<<k))>>k);
			cout << endl;
			return 0;
		}
	}
}
```