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 a[620][620], n, d1, d2, z;
void gao(int d) {
	int b = 0;
	while (d % 4 == 0) {
		d /= 4;
		b++;
	}
	if (d & 1) {
		for (int i = 0; i < 2 * n; i++) {
			for (int j = 0; j < 2 * n; j++) {
				if (((i >> b) + (j >> b)) & 1) {
					a[i][j] = 1;
				}
			}
		}
	} else {
		for (int i = 0; i < 2 * n; i++) {
			for (int j = 0; j < 2 * n; j++) {
				if ((i >> b) & 1) {
					a[i][j] = 1;
				}
			}
		}
	}
}
int main() {
	scanf("%d%d%d", &n, &d1, &d2);
	gao(d1);
	gao(d2);
	for (int i = 0; i < 2 * n; i++) {
		for (int j = 0; j < 2 * n; j++) {
			if (z < n * n && !a[i][j]) {
				printf("%d %d\n", i, j);
				z++;
			}
		}
	}
	return 0;
}