Takahashi is doing a research on sets of points in a plane. Takahashi thinks a set S of points in a coordinate plane is a good set when S satisfies both of the following conditions:

* The distance between any two points in S is not \sqrt{D_1}.
* The distance between any two points in S is not \sqrt{D_2}.



Here, D_1 and D_2 are positive integer constants that Takahashi specified.

Let X be a set of points (i,j) on a coordinate plane where i and j are integers and satisfy 0 ≤ i,j < 2N.

Takahashi has proved that, for any choice of D_1 and D_2, there exists a way to choose N^2 points from X so that the chosen points form a good set. However, he does not know the specific way to choose such points to form a good set. Find a subset of X whose size is N^2 that forms a good set.

Constraints

* 1 ≤ N ≤ 300
* 1 ≤ D_1 ≤ 2×10^5
* 1 ≤ D_2 ≤ 2×10^5
* All values in the input are integers.

Input

Input is given from Standard Input in the following format:


N D_1 D_2


Output

Print N^2 distinct points that satisfy the condition in the following format:


x_1 y_1
x_2 y_2
:
x_{N^2} y_{N^2}


Here, (x_i,y_i) represents the i-th chosen point. 0 ≤ x_i,y_i < 2N must hold, and they must be integers. The chosen points may be printed in any order. In case there are multiple possible solutions, you can output any.

Examples

Input

2 1 2


Output

0 0
0 2
2 0
2 2


Input

3 1 5


Output

0 0
0 2
0 4
1 1
1 3
1 5
2 0
2 2
2 4
#include <bits/stdc++.h>

using namespace std;

int n, d1, d2;

vector<pair<int, int>> salam[4];

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> d1 >> d2;
	int L1 = 0;
	while (!(d1 & 1))
	{
		L1++;
		d1 >>= 1;
	}
	int L2 = 0;
	while (!(d2 & 1))
	{
		L2++;
		d2 >>= 1;
	}

	for (int i = 0; i < 2 * n; i++)
	{
		for (int j = 0; j < 2 * n; j++)
		{
			int color = 0;
			if (L1 & 1) {
				int l1 = L1 / 2;
				if (i & (1 << l1))
					color += 1;
			}
			else {
				int l1 = L1 / 2;
				if ((i & (1 << l1)) ^ (j & (1 << l1)))
					color += 1;
			}
			if (L2 & 1) {
				int l2 = L2 / 2;
				if (i & (1 << l2))
					color += 2;
			}
			else {
				int l2 = L2 / 2;
				if ((i & (1 << l2)) ^ (j & (1 << l2)))
					color += 2;
			}
			salam[color].push_back({i, j});
		}
	}
	for (int i = 0; i < 4; i++)
		if ((int)salam[i].size() >= (n * n))
		{
			for (int j = 0; j < n * n; j++)
			{
				cout << salam[i][j].first << ' ' << salam[i][j].second << '\n';
			}
			break;
		}
	return 0;
}