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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

int dp[21][2100000];
int anslen,ans;
int n,k;
int store[2100000];

void upd(int tmp,int ori,int len)
{
	store[ori | (1 << len)] += tmp;
	if(store[ori | (1 << len)] >= k)
	{
		if(len > anslen) ans = ori,anslen = len;
		else if(len == anslen) ans = min(ans,ori);
	}
}

char str[1050000];

int fuckfir1(int val)
{
	return val ^ (1 << (31 - __builtin_clz(val)));
}

int rem0len[1050000];
int rem0cur[1050000];
bool mark[1050000];

void print(int val,int lef)
{
	if(!lef) return ;
	if(lef == 1)
	{
		assert(val <= 1);
		printf("%d",val % 2);
		return ;
	}
	print(val / 2,lef - 1);
	printf("%d",val % 2);
}

int main()
{
	for(int i = 2;i <= 1048576;i *= 2) mark[i - 1] = 1;
//	printf("%d\n",__builtin_clz(2));
	ans = -1;
	scanf("%d%d",&n,&k);
	for(int i = 0;i <= n; ++ i)
	{
		int ta = (1 << i);
		scanf("%s",str);
		for(int j = 0;j < (1 << i); ++ j)
			dp[i][ta | j] = str[j] - '0';
	}
	for(int i = (1 << n) - 1;i >= 0; -- i)
	{
		if(mark[i]) continue;
		int cur = i;
		bool flg1 = 0;
		for(int j = n;j >= 0; -- j)
		{
			if(!(cur & 1 << j - 1) && flg1)
			{
				rem0len[i] = j - 1;
				rem0cur[i] = cur;
				break;
			}
			if(cur & 1 << j - 1) flg1 = 1,cur ^= 1 << j - 1;
		}
	}
	int all = (1 << n + 1) - 1;
	for(int i = n;i >= 0; -- i)
	{
		for(int j = all;j >= 0; -- j)
		{
			int cur = j & ((1 << i) - 1),lef = j >> i;
			int len = 31 - __builtin_clz(lef);
			upd(dp[i][j],fuckfir1(lef),len);
			if(!dp[i][j]) continue;
			if(cur)
			{
				int newlef = lef * 2 + 1;
				int newlen = (31 - __builtin_clz(cur));
				int newcur = fuckfir1(cur);
		//		printf("%d %d %d +1 -> %d %d %d\n",cur,fuckfir1(lef),i,newcur,fuckfir1(newlef),newlen);
				dp[newlen][newlef << newlen | newcur] += dp[i][j];
			}
			if(cur != (1 << i) - 1)
			{
				int newlef = lef * 2;
				if(cur & (1 << i - 1))
				{
					int newlen = rem0len[cur];
					int newcur = rem0cur[cur];
		//			printf("%d %d %d +0 -> %d %d %d\n",cur,fuckfir1(lef),i,newcur,fuckfir1(newlef),newlen);
					dp[newlen][newlef << newlen | newcur] += dp[i][j];
				}
				else
				{
					int newlen = i - 1;
					int newcur = cur;
		//			printf("%d %d %d +0 -> %d %d %d\n",cur,fuckfir1(lef),i,newcur,fuckfir1(newlef),newlen);
					dp[newlen][newlef << newlen | newcur] += dp[i][j];
				}
			}
		}
	}
	if(ans == -1) puts("");
	else
	{
		print(ans,anslen);
	}
}