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[360005], fa[360005];
bool d[360005];
vector<pair<int, int>> ans[4];

int Sqrt(int x) {
	int y = sqrt(x);
	while (y * y <= x) ++y;
	while (y * y > x) --y;
	return y;
}

int find(int x) {
	if (fa[x] == x) return x;
	int ff = find(fa[x]);
	d[x] ^= d[fa[x]];
	return fa[x] = ff;
}
void merge(int x, int y) {
	int s = find(x), t = find(y);
	if (s == t) return;
	fa[s] = t, d[s] = d[x] ^ d[y] ^ 1;
}

int main() {
	scanf("%d", &n);
	for (int c = 0; c < 2; ++c) {
		scanf("%d", D + c);
		for (int i = 0; i < 4 * n * n; ++i) fa[i] = i, d[i] = 0;
		for (int i = 0; i * i <= D[c]; ++i) {
			int j = Sqrt(D[c] - i * i);
			if (i * i + j * j != D[c]) continue;
			for (int a = 0; a + i < 2 * n; ++a)
				for (int b = 0; b + j < 2 * n; ++b) {
					merge(a * 2 * n + b, (a + i) * 2 * n + b + j);
					merge((a + i) * 2 * n + b, a * 2 * n + b + j);
				}
		}
		for (int i = 0; i < 4 * n * n; ++i)
			find(i), col[i] |= d[i] << c;
	}
	for (int i = 0; i < 2 * n; ++i)
		for (int j = 0; j < 2 * n; ++j)
			ans[col[i * 2 * n + j]].emplace_back(i, j);
	for (int c = 0; c < 4; ++c)
		if (ans[c].size() >= n * n) {
			for (int i = 0; i < n * n; ++i)
				printf("%d %d\n", ans[c][i].first, ans[c][i].second);
			return 0;
		}
}
