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>
#define Rint register int
using namespace std;
const int N = 603;
int n, D1, D2, cnt;
bool vis[N][N];
inline void calc(int d){
	int len = 0;
	while(!(d & 3)) d /= 4, ++ len;
	if(d & 1)
		for(Rint i = 0;i < n * 2;i ++)
			for(Rint j = 0;j < n * 2;j ++)
				vis[i][j] |= (((i >> len) & 1) ^ ((j >> len) & 1));
	else if(!(d & 1))
		for(Rint i = 0;i < n * 2;i ++)
			for(Rint j = 0;j < n * 2;j ++)
				vis[i][j] |= ((i >> len) & 1);
}
int main(){
	scanf("%d%d%d", &n, &D1, &D2);
	calc(D1); calc(D2);
	for(Rint i = 0;i < 2 * n && cnt < n * n;i ++)
		for(Rint j = 0;j < 2 * n && cnt < n * n;j ++)
			if(!vis[i][j]){++ cnt; printf("%d %d\n", i, j);}
}