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;

namespace TYC
{
	const int N = 605;

	int n, vis[N][N], cnt[5];

	void solve(int D, const int type)
	{
		int tim = 0;
		while (!(D & 3))
			D /= 4, tim++;
		if (D & 1)
		{
			for (int i = 0; i < n; i++)
				for (int j = 0; j < n; j++)
					if (((i >> tim) + (j >> tim)) & 1)
						vis[i][j] += type;
		}
		else
		{
			for (int i = 0; i < n; i++)
				for (int j = 0; j < n; j++)
					if ((i >> tim) & 1)
						vis[i][j] += type;
		}
	}

	void work()
	{
		int d1, d2;
		scanf("%d%d%d", &n, &d1, &d2);
		n <<= 1;
		solve(d1, 1);
		solve(d2, 2);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				cnt[vis[i][j]]++;
		int mx = 0, id = -1;
		for (int i = 0; i < 4; i++)
			if (cnt[i] > mx)
				mx = cnt[i], id = i;
		for (int i = 0, res = (n >> 1) * (n >> 1); i < n; i++)
			for (int j = 0; j < n && res; j++)
				if (vis[i][j] == id)
				{
					printf("%d %d\n", i, j);
					res--;
				}
	}
}

int main()
{
	TYC::work();
	return 0;
}