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, d1, d2, n2, vis[600][600], p;
void run(int d) {
	int b = 0;
	while (!(d & 3)) d >>= 2, b++;
	if (d & 1) {
		for (int i = 0; i < n2; i++)
			for (int j = 0; j < n2; j++) 
				if (((i >> b) + (j >> b)) & 1) vis[i][j] = 1;
	} else {
		for (int i = 0; i < n2; i++)
			for (int j = 0; j < n2; j++) 
				if ((i >> b) & 1) vis[i][j] = 1;
	}
}
int main() {
	scanf("%d%d%d", &n, &d1, &d2); n2 = n * 2;
	run(d1), run(d2);	
	for (int i = 0; i < n2; i++)
		for (int j = 0; j < n2; j++)
			if (p < n * n && !vis[i][j])
				printf("%d %d\n", i, j), p++;
//	for (int i = 0; i < n2; i++, puts(""))
//		for (int j = 0; j < n2; j++)
//			printf("%d", vis[i][j]);
	return 0;
} 