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 eps 1e-7
using namespace std;
int n,d1,d2,cnt=0;
bool f[1005][1005];
inline void solve(int d){
	int b=0;
	while(!(d%4))d>>=2,++b;
	if(d&1){
		for(int i=0;i<(n<<1);++i)
			for(int j=0;j<(n<<1);++j)
				if(((i>>b)+(j>>b))&1)f[i][j]=true;
		return;
	}
	for(int i=0;i<(n<<1);++i)
		for(int j=0;j<(n<<1);++j)
			if((i>>b)&1)f[i][j]=true;
}
int main(){
	scanf("%d%d%d",&n,&d1,&d2);
	solve(d1),solve(d2);
	for(int i=0;i<(n<<1);++i)
		for(int j=0;j<(n<<1);++j)
			if(cnt<n*n&&!f[i][j])printf("%d %d\n",i,j),++cnt;
	return 0;
}