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 <cstdio>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <vector>
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define sz(a) ((int) (a).size())

const int N = 300;
int color[2][2 * N][2 * N], n, d[2];
vector< pair<int, int> > dirs[2], all[2][2];

void dfs(int c, int x, int y, int cur) {
	color[c][x][y] = cur;
	for (int i = 0; i < sz(dirs[c]); ++i) {
		int nx = x + dirs[c][i].fs, ny = y + dirs[c][i].sc;
		if ((nx >= 0) && (ny >= 0) && (nx < 2 * n) && (ny < 2 * n) && (color[c][nx][ny] == -1)) {
			dfs(c, nx, ny, cur ^ 1);
		}
	}
}

int main() {
	scanf("%d%d%d", &n, &d[0], &d[1]);
	for (int c = 0; c < 2; ++c) {
		for (int dx = -2 * n + 1; dx < 2 * n; ++dx) {
			for (int dy = -2 * n + 1; dy < 2 * n; ++dy) {
				int cd = dx * dx + dy * dy;
				if (cd == d[c]) {
					dirs[c].pb(mp(dx, dy));
				}
			}
		}
	}
	memset(color, -1, sizeof(color));
	for (int c = 0; c < 2; ++c) {
		for (int i = 0; i < 2 * n; ++i) {
			for (int j = 0; j < 2 * n; ++j) {
				if (color[c][i][j] == -1) {
					dfs(c, i, j, 0);
				}
			}
		}
	}
	for (int i = 0; i < 2 * n; ++i) {
		for (int j = 0; j < 2 * n; ++j) {
			vector< pair<int, int> > &who = all[color[0][i][j]][color[1][i][j]];
			who.pb(mp(i, j));
			if (sz(who) == n * n) {
				for (int k = 0; k < n * n; ++k) {
					printf("%d %d\n", who[k].fs, who[k].sc);
				}
				return 0;
			}
		}
	}
	return 0;
}