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;
const int N=666*666;
int n,d1,d2,vis[N],ans[N];
vector<int>e[N],qwq[4];
int hx(int a,int b){return a*n+b;}
void add(int a,int b){e[a].push_back(b),e[b].push_back(a);}
void dfs(int v,int f,int x){
	if(vis[v])return;vis[v]=1;if(x)ans[v]|=f;
	for(int i=0;i<(int)e[v].size();i++)dfs(e[v][i],f,x^1);
}
void solve(int d,int f){
	memset(vis,0,sizeof(vis));
	for(int i=0;i<n*n;i++)e[i].clear();
	for(int i=0;i*i<=d;i++){
		int j=sqrt(d-i*i);
		if(i*i+j*j==d){
			for(int x=0;x<n;x++)for(int y=0;y<n;y++){
				int X=x+i,Y=y+j;
				if(0<=X&&X<n&&0<=Y&&Y<n)add(hx(x,y),hx(X,Y));
				Y=y-j;
				if(0<=X&&X<n&&0<=Y&&Y<n)add(hx(x,y),hx(X,Y));
			}
		}
	}
	for(int i=0;i<n*n;i++)if(!vis[i])dfs(i,f,0);
}
int main(){
    scanf("%d%d%d",&n,&d1,&d2);n<<=1;
	solve(d1,1);
	solve(d2,2);
	//for(int i=0;i<n;i++){for(int j=0;j<n;j++)printf("%d ",ans[hx(i,j)]);printf("\n");}
	for(int i=0;i<n;i++)for(int j=0;j<n;j++)qwq[ans[hx(i,j)]].push_back(hx(i,j));
	for(int i=0;i<4;i++)if((int)qwq[i].size()>=n*n/4){
		for(int j=0;j<n*n/4;j++)printf("%d %d\n",qwq[i][j]/n,qwq[i][j]%n);
		return 0;
	}
    return 0;
}