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 fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
#define P pair<int,int>
#define fi first
#define se second
#define pb push_back
using namespace std;
const int N=605;
int n,d1,d2;
int c[N][N],su[9];
bool vi[N][N];
vector<P>b[3];
void get(int x,vector<P>&b){
	fo(i,0,n*2){
		int a=x-i*i;
		if(x<0)break;
		int c=sqrt(a);
		if(c*c==a){
			b.pb(P(i,c));
			b.pb(P(i,-c));
			b.pb(P(-i,c));
			b.pb(P(-i,-c));
		}
	}
}
void go(int x,int y,int o,int o2){
	vi[x][y]=1;
	c[x][y]|=o2;
	for(P i:b[o]){
		int xx=x+i.fi,yy=y+i.se;
		if(xx>=0&&yy>=0&&xx<n*2&&yy<n*2&&!vi[xx][yy])
			go(xx,yy,o,o-o2);
	}
}
int main(){
	cin>>n>>d1>>d2;
	get(d1,b[1]);
	get(d2,b[2]);
	fo(i,1,2){
		memset(vi,0,sizeof vi);
		fo(j,0,n*2-1)
			fo(k,0,n*2-1)
				if(!vi[j][k])go(j,k,i,0);
	}
	fo(j,0,n*2-1)
		fo(k,0,n*2-1)
			su[c[j][k]]++;
	int mx=0,cnt=0;
	fo(i,1,3)if(su[i]>su[mx])mx=i;
	fo(j,0,n*2-1)
		fo(k,0,n*2-1)
			if(c[j][k]==mx){
				printf("%d %d\n",j,k);
				if(++cnt==n*n)return 0;
			}
}