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<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm>

using namespace std;

#define LL long long
#define fgx cerr<<"--------------"<<endl;
#define dgx cerr<<"=============="<<endl;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int MAXN = 100010;
const int INF = 2147483600;

int N,D1,D2;
bool Vis[710][710];

inline void solve(int P){
	int K=0; while(P%4==0) P>>=2,K++;
	if(P%4==3) return ;
	if(P%4==1) {
		for(int i=0;i<N;i++)
			for(int j=0;j<N;j++)
				Vis[i][j]&=((((i>>K)^(j>>K))&1)^1);
	} else{
		for(int i=0;i<N;i++)
			for(int j=0;j<N;j++) 
				Vis[i][j]&=(((i>>K)&1)^1);
	} return ;
}

int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	N=read(),D1=read(),D2=read();
	int cnt=N*N;
	N<<=1; memset(Vis,1,sizeof(Vis));
	solve(D1); solve(D2); 
	for(int i=0;i<N;i++){
		for(int j=0;j<N;j++){
			if(Vis[i][j]){
				printf("%d %d\n",i,j); --cnt;
				if(!cnt) return 0;
			}
		}
	}
	return 0;
}


