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, d[2], col[2][360007];
vector<pair<int, int> >e[2];
int id(int a, int b) { return a * 2 * n + b; }
int check(int pos) { return pos >= 0 && pos < 2 * n; }
int dfs(int p, int a, int b)
{
	for (int i = 0, lim = e[p].size(); i < lim; i++)
	{
		int c = a + e[p][i].first, d = b + e[p][i].second;
		if (!check(c) || !check(d))continue;
		if (!col[p][id(c, d)])
		{
			col[p][id(c, d)] = -col[p][id(a, b)];
			dfs(p, c, d);
		}
	}
	return 0;
}
int main()
{
	cin >> n >> d[0] >> d[1];
	for (int p = 0; p <= 1; p++)
		for (int i = 0; i <= d[p]; i++)
		{
			int j = sqrt(d[p] - i * i);
			if (i * i + j * j != d[p])continue;
			e[p].push_back(make_pair(i, j));
			e[p].push_back(make_pair(-i, j));
			e[p].push_back(make_pair(i, -j));
			e[p].push_back(make_pair(-i, -j));
		}
	for (int i = 0; i < 2 * n; i++)
		for (int j = 0; j < 2 * n; j++)
			for (int p = 0; p <= 1; p++)
				if (!col[p][id(i, j)])
				{
					col[p][id(i, j)] = 1;
					dfs(p, i, j);
				}
	int cnt = 0;
	for (int i = 0; i < 2 * n; i++)
		for (int j = 0; j < 2 * n; j++)
			if (col[0][id(i, j)] == 1 && col[1][id(i, j)] == 1)
			{
				printf("%d %d\n", i, j);
				if (++cnt == n * n)return 0;
			}
	return 0;
}