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>
int arr[605][605], app[5]; 
void work(int n, int d, int mask)
{
	int cnt = 0; 
	while (!(d & 3))
	{
		d >>= 2; 
		cnt++; 
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			if (((i >> cnt) ^ (d & 1 ? j >> cnt : 0)) & 1)
				arr[i][j] |= mask;
		}
	}
}
int main()
{
	// freopen("AGC025-D.in", "r", stdin); 
	int n, d1, d2;
	scanf("%d%d%d", &n, &d1, &d2); 
	work(n << 1, d1, 1); 
	work(n << 1, d2, 2); 
	for (int i = 0; i < n << 1; i++)
	{
		for (int j = 0; j < n << 1; j++)
			app[arr[i][j]]++; 
	}
	int val = 0, rem = n * n; 
	for (int i = 0; i < 4; i++)
	{
		if (app[i] >= rem)
			val = i; 
	}
	for (int i = 0; i < n << 1; i++)
	{
		for (int j = 0; j < n << 1; j++)
		{
			if (rem && arr[i][j] == val)
			{
				printf("%d %d\n", i, j);
				rem--;
			}
		}
	}
	return 0; 
}
