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;
typedef long long ll;

const int MAXN = (1LL<<21);

int n, k, conv[MAXN], nex[MAXN][2];

string find(int m, vector<int> ct) {
	assert(ct.size() == (1<<(m+1)));
	int sum = 0;
	for (int t : ct)
		sum += t;
	if (sum < k)
		return "a";
	if (m == 0)
		return "";

	string best = "";
	vector<int> v;
	v.resize(1<<m);

	for (int i = 1; i < ct.size(); ++i)
		if (nex[i][0] != -1)
			v[nex[i][0]] += ct[i];

	string ans0 = find(m-1, v);
	if (ans0 != "a") {
		ans0 = "0" + ans0;
		if (best.length() < ans0.length())
			best = ans0;
		else if (best.length() == ans0.length())
			best = min(best, ans0);
	}

	v.clear();
	v.resize(1<<m);

	for (int i = 1; i < ct.size(); ++i)
		if (nex[i][1] != -1)
			v[nex[i][1]] += ct[i];

	string ans1 = find(m-1, v);
	if (ans1 != "a") {
		ans1 = "1" + ans1;
		if (best.length() < ans1.length())
			best = ans1;
		else if (best.length() == ans1.length())
			best = min(best, ans1);
	}

	return best;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n >> k;
	vector<int> v;
	v.resize(1<<(n+1));

	for (int i = 0; i <= n; ++i) {
		string g;
		cin >> g;
		assert(g.size() == (1<<i));
		for (int j = 0; j < (1<<i); ++j)
			if (g[j] == '1')
				v[(1<<i) + j]++;
	}

	for (int l = 0; l <= n; ++l)
		for (int j = 0; j < (1<<l); ++j) {
			if (l == 0) {
				nex[(1<<l)+j][0] = nex[(1<<l)+j][1] = -1;
				continue;
			}

			if (j/(1<<(l-1)) == 0) {
				nex[(1<<l)+j][0] = (1<<(l-1))+j;
				nex[(1<<l)+j][1] = nex[(1<<(l-1))+j][1];
			} else {
				nex[(1<<l)+j][0] = nex[j][0];
				nex[(1<<l)+j][1] = j;
			}
		}

	cout << find(n, v) << endl;
}