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 title "title"
#define ll long long
#define ull unsigned ll
#define fix(x) fixed<<setprecision(x)
#define pii pair<int,int>
#define vint vector<int>
#define pb push_back
using namespace std;
void Freopen(){
	freopen(title".in","r",stdin);
	freopen(title".out","w",stdout);
}
int read(){
	int g=0,f=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch&&ch<='9'){g=g*10+ch-'0';ch=getchar();}
	return g*f;
}
const int N=605;
bool vis[N][N];
int n,d1,d2,cnt;
void solve(int dis){
	int tim=0;
	while(!(dis%4))dis/=4,tim++;
	if(dis&1){for(int i=0;i<n<<1;i++)for(int j=0;j<n<<1;j++)if((i>>tim)+(j>>tim)&1)vis[i][j]=true;}
	else{for(int i=0;i<n<<1;i++)for(int j=0;j<n<<1;j++)if((i>>tim)&1)vis[i][j]=true;}
}
signed main(){
	n=read(),d1=read(),d2=read();
	solve(d1),solve(d2);
	for(int i=0;i<n<<1;i++)for(int j=0;j<n<<1;j++){
		if(!vis[i][j])cout<<i<<' '<<j<<'\n',cnt++;
		if(cnt==n*n)return signed();
	}return signed();
}