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>
#include <cstring>
#include <algorithm>
int n,n2,d1,d2,c1 = 0,c2 = 0,cnt = 0,tot;
int main()
{
	scanf("%d%d%d",&n,&d1,&d2);n2 = n << 1,tot = n * n;
	while(!(d1 % (1 << (c1 + 1)))) ++c1;
	while(!(d2 % (1 << (c2 + 1)))) ++c2;
	for(int i = 0;i < n2 && cnt < tot;i++) for(int j = 0;j < n2 && cnt < tot;j++)
	{
		int b1 = (c1 & 1) ? (!((i >> (c1 >> 1)) & 1)) : (((i >> (c1 >> 1)) & 1) == ((j >> (c1 >> 1)) & 1));
		int b2 = (c2 & 1) ? (!((i >> (c2 >> 1)) & 1)) : (((i >> (c2 >> 1)) & 1) == ((j >> (c2 >> 1)) & 1));
		if(b1 && b2 && (++cnt)) printf("%d %d\n",i,j);
	}
	return 0;
}
