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;
const int maxn = 4e5 + 10;
vector<int> E[2][maxn]; 
int col[2][maxn];
int cnt[4];
int n, m, d1, d2;
bool in(int x, int y) { return 0 <= x && x < (n << 1) && 0 <= y && y < (n << 1);}
int idx(int x, int y) { return x * (n << 1) + y;}
void connect(int x, int y, int k)
{
	for(int i = 0; i < (n << 1); ++i)
		for(int j = 0; j < (n << 1); ++j)
			if(in(i + x, j + y))
			{
				int u = idx(i, j), v = idx(i + x, j + y);
				E[k][u].push_back(v);
				E[k][v].push_back(u);
			}
}
void dfs(int x, int k, int c)
{
	col[k][x] = c;
	for(auto v : E[k][x])
		if(col[k][v] == -1)
			dfs(v, k, c ^ 1);
}
void solve(int d, int k)
{
	int a = sqrt(d) + 0.5;
	for(int i = -a; i <= a; ++i)
	{
		if(i * i > d) continue;
		int j = sqrt(d - i * i) + 0.5;
		if(i * i + j * j == d)
			connect(i, j, k);
	}
	memset(col[k], -1, sizeof col[k]);
	for(int i = 0; i < m; ++i)
		if(col[k][i] == -1)
			dfs(i, k, 0);
}
int main()
{
	scanf("%d%d%d", &n, &d1, &d2);
	m = 4 * n * n;
	solve(d1, 0);
	solve(d2, 1);
	for(int i = 0; i < m; ++i)
		++cnt[col[0][i] * 2 + col[1][i]];
	int val = max_element(cnt, cnt + 4) - cnt;
	for(int i = 0, tot = n * n; tot; ++i)
		if(col[0][i] * 2 + col[1][i] == val)
		{
			--tot;
			printf("%d %d\n", i / (n << 1), i % (n << 1));
		}
	return 0;
}