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>
inline void work(int n, int d, int arr[][605])
{
	int blk = 1; 
	while (!(d & 3))
	{
		d >>= 2; 
		blk <<= 1; 
	}
	if (d & 1)
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
				arr[i][j] = (i / blk & 1) ^ (j / blk & 1);
		}
	}
	else
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
				arr[i][j] = i / blk & 1;
		}
	}
}
int arr1[605][605], arr2[605][605], app[5]; 
int main()
{
	// freopen("AGC025-D.in", "r", stdin); 
	int n, d1, d2; 
	scanf("%d%d%d", &n, &d1, &d2);
	work(n << 1, d1, arr1); 
	work(n << 1, d2, arr2); 
	for (int i = 0; i < n << 1; i++)
	{
		for (int j = 0; j < n << 1; j++)
			app[arr1[i][j] << 1 | arr2[i][j]]++; 
	}
	for (int i = 0; i < 4; i++)
	{
		if (app[i] < n * n)
			continue;
		int cnt = n * n; 
		for (int x = 0; x < n << 1; x++)
		{
			for (int y = 0; y < n << 1; y++)
			{
				if ((arr1[x][y] << 1 | arr2[x][y]) == i && cnt)
				{
					printf("%d %d\n", x, y);
					cnt--;
				}
			}
		}
		break; 
	}
	return 0; 
}
