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>
using namespace std;
inline int read(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 605
pair<int,int> mp[MN<<1];
int N,D1,D2,cc;
bool col[MN][MN][2];
struct node{int x,y;};
vector<node> G[5];
void solve(int D,int l){
	register int i,j,k,ct=0,squ=1;
	while(~D&1){ct++;D>>=1;}
	squ=1<<(ct>>1);
	if(~ct&1){
		for(i=0;i<N*2;i++)for(j=0;j<N*2;j++)col[i][j][l]=((i/squ)+(j/squ))&1;
	}
	else{
		for(i=0;i<N*2;i++)for(j=0;j<N*2;j++) col[i][j][l]=(i/squ)&1;
	}
}
int main(){
	register int i,j;
	N=read(),D1=read(),D2=read();
	solve(D1,0),solve(D2,1);
	for(i=0;i<N<<1;i++)for(j=0;j<N<<1;j++)
	G[col[i][j][0]+col[i][j][1]<<1].push_back((node){i,j});
	for(i=0;i<4;i++)if(G[i].size()>=N*N){
		for(j=0;j<N*N;j++) printf("%d %d\n",G[i][j].x,G[i][j].y);
		return 0;
	}
	return 0;
}