#include <iostream>
#include <cstdio>
using namespace std;
int ch[2][1 << 22], id[22][1 << 22], cnt, a[22][1 << 22], siz[22], c[22][1 << 22], num[22][1 << 22], ans[22], n, k;
char s[1 << 22];
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void get()
{
	id[0][0] = (cnt ++); ch[0][id[0][0]] = ch[1][id[0][0]] = -1;
	for(int i = 1; i <= 20; i ++)
	{
		int t = (1 << i), k = (t >> 1);
		for(int s = 0; s < t; s ++)
		{
			id[i][s] = (cnt ++); int p = (s & k), q = (p ? 1 : 0);
			ch[q][id[i][s]] = id[i - 1][s ^ p];
			ch[!q][id[i][s]] = ch[!q][id[i - 1][s ^ p]];
		}
	}
}
void dfs(int d, int s)
{
	if(ans[d] == -1) ans[d] = s;
	for(int p = 0; p <= 1; p ++)
	{
		int tot = 0;
		for(int i = 0; i < siz[d]; i ++)
		{
			int to = ch[p][a[d][i]]; if(to == -1) continue;
			if(num[d + 1][to] == -1) num[d + 1][a[d + 1][siz[d + 1]] = to] = siz[d + 1], siz[d + 1] ++;
			tot += c[d][i]; c[d + 1][num[d + 1][to]] += c[d][i];
		}
		if(tot >= k) dfs(d + 1, (s << 1) | p);
		for(int i = 0; i < siz[d + 1]; i ++) num[d + 1][a[d + 1][i]] = -1, c[d + 1][i] = 0;
		siz[d + 1] = 0;
	}
}
int main()
{
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	n = read(); k = read(); get();
	for(int i = 0; i <= n; i ++) for(int j = 0; j < cnt; j ++) num[i][j] = -1;
	for(int i = 0; i <= n; i ++)
	{
		scanf("%s", s); int t = (1 << i);
		for(int j = 0; j < t; j ++) if(s[j] == '1') num[0][a[0][siz[0]] = id[i][j]] = siz[0], c[0][num[0][id[i][j]]] ++, siz[0] ++;
		ans[i] = -1;
	}
	dfs(0, 0);
	for(int i = n; i >= 0; i --)
		if(ans[i] != -1)
		{
			for(int j = i - 1; j >= 0; j --)
			putchar(((ans[i] >> j) & 1) + '0');
			puts(""); return 0;
		}
	fclose(stdin);
	fclose(stdout);
	return 0;
}

### Prompt

Your task is to create a CPP solution 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>
using namespace std;
int ch[2][1 << 22], id[22][1 << 22], cnt, a[22][1 << 22], siz[22], c[22][1 << 22], num[22][1 << 22], ans[22], n, k;
char s[1 << 22];
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void get()
{
	id[0][0] = (cnt ++); ch[0][id[0][0]] = ch[1][id[0][0]] = -1;
	for(int i = 1; i <= 20; i ++)
	{
		int t = (1 << i), k = (t >> 1);
		for(int s = 0; s < t; s ++)
		{
			id[i][s] = (cnt ++); int p = (s & k), q = (p ? 1 : 0);
			ch[q][id[i][s]] = id[i - 1][s ^ p];
			ch[!q][id[i][s]] = ch[!q][id[i - 1][s ^ p]];
		}
	}
}
void dfs(int d, int s)
{
	if(ans[d] == -1) ans[d] = s;
	for(int p = 0; p <= 1; p ++)
	{
		int tot = 0;
		for(int i = 0; i < siz[d]; i ++)
		{
			int to = ch[p][a[d][i]]; if(to == -1) continue;
			if(num[d + 1][to] == -1) num[d + 1][a[d + 1][siz[d + 1]] = to] = siz[d + 1], siz[d + 1] ++;
			tot += c[d][i]; c[d + 1][num[d + 1][to]] += c[d][i];
		}
		if(tot >= k) dfs(d + 1, (s << 1) | p);
		for(int i = 0; i < siz[d + 1]; i ++) num[d + 1][a[d + 1][i]] = -1, c[d + 1][i] = 0;
		siz[d + 1] = 0;
	}
}
int main()
{
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	n = read(); k = read(); get();
	for(int i = 0; i <= n; i ++) for(int j = 0; j < cnt; j ++) num[i][j] = -1;
	for(int i = 0; i <= n; i ++)
	{
		scanf("%s", s); int t = (1 << i);
		for(int j = 0; j < t; j ++) if(s[j] == '1') num[0][a[0][siz[0]] = id[i][j]] = siz[0], c[0][num[0][id[i][j]]] ++, siz[0] ++;
		ans[i] = -1;
	}
	dfs(0, 0);
	for(int i = n; i >= 0; i --)
		if(ans[i] != -1)
		{
			for(int j = i - 1; j >= 0; j --)
			putchar(((ans[i] >> j) & 1) + '0');
			puts(""); return 0;
		}
	fclose(stdin);
	fclose(stdout);
	return 0;
}

```