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 <stdio.h>
int n, a, b, x, y;
int z[600][600];
int w[600][600];
int dx[8];
int dy[8];
int main() {
	scanf("%d%d%d", &n, &a, &b);
	while (!(a % 2)) {
		x++;
		a /= 2;
	}
	while (!(b % 2)) {
		y++;
		b /= 2;
	}
	a = 1 << (x / 2);
	b = 1 << (y / 2);
	if (x % 2) {
		dx[0] = a; dy[0] = a;
		dx[1] = -a; dy[1] = a;
		dx[2] = a; dy[2] = -a;
		dx[3] = -a; dy[3] = -a;
	}
	else {
		dx[0] = a;
		dy[1] = a;
		dx[2] = -a;
		dy[3] = -a;
	}
	if (y % 2) {
		dx[4] = b; dy[4] = b;
		dx[5] = -b; dy[5] = b;
		dx[6] = b; dy[6] = -b;
		dx[7] = -b; dy[7] = -b;
	}
	else {
		dx[4] = b;
		dy[5] = b;
		dx[6] = -b;
		dy[7] = -b;
	}
	for (int i = 0; i < 2 * n; i++) {
		for (int j = 0; j < 2 * n; j++) {
			if (!z[i][j]) {
				for (int k = 0; k < 4; k++) {
					x = i + dx[k];
					y = j + dy[k];
					if (0 <= x && x < 2 * n && 0 <= y && y < 2 * n) {
						z[x][y] = 1;
					}
				}
			}
			if (!w[i][j]) {
				for (int k = 4; k < 8; k++) {
					x = i + dx[k];
					y = j + dy[k];
					if (0 <= x && x < 2 * n && 0 <= y && y < 2 * n) {
						w[x][y] = 1;
					}
				}
			}
		}
	}
	x = 0;
	for (int i = 0; i < 2 * n; i++) {
		for (int j = 0; j < 2 * n; j++) {
			if (!(z[i][j] | w[i][j])) {
				x++;
				printf("%d %d\n", i, j);
				if (x == n * n)return 0;
			}
		}
	}
}