Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

2 1 2

Sample Output 1

0 0
0 2
2 0
2 2

Among these points, the distance between 2 points is either 2 or 2\sqrt{2}, thus it satisfies the condition.

Sample Input 2

3 1 5

Sample Output 2

0 0
0 2
0 4
1 1
1 3
1 5
2 0
2 2
2 4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define SZ(x) ((int)x.size())
#define L(i,u) for (register int i=head[u]; i; i=nxt[i])
#define rep(i,a,b) for (register int i=(a); i<=(b); i++)
#define per(i,a,b) for (register int i=(a); i>=(b); i--)
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef pair<int,int> Pii;
typedef pair<Pii,Pii> Ppp;
typedef vector<int> Vi;
inline void read(int &x) {
	x=0; char c=getchar(); int f=1;
	while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
	while (isdigit(c)) {x=x*10+c-'0'; c=getchar();} x*=f;
}
inline ui R() {
	static ui seed=416;
	return seed^=seed>>5,seed^=seed<<17,seed^=seed>>13;
}
const int N = 606;
int n,d1,d2;vector<Pii> res;
void solve(int d1, int d2, vector<Ppp> s){//now,real
	vector<Ppp> a,b;
	rep(i,0,SZ(s)-1)if(!((s[i].fi.fi+s[i].fi.se)&1)){
		a.pb(Ppp(Pii((s[i].fi.fi+s[i].fi.se)/2,(s[i].fi.se-s[i].fi.fi)/2),s[i].se));
	}
	else{
		s[i].fi.fi++;
		b.pb(Ppp(Pii((s[i].fi.fi+s[i].fi.se)/2,(s[i].fi.se-s[i].fi.fi)/2),s[i].se));
	}
	if(SZ(a)<SZ(b))swap(a,b);
	if(d1&d2&1)rep(i,0,SZ(a)-1)res.pb(a[i].se);
	else{
		if(d2&1)d2=d1;if(d1&1)d1=d2;
		solve(d1/2,d2/2,a);solve(d1/2,d2/2,b);
	}
}
int main() {
	read(n);read(d1);read(d2);
	vector<Ppp>d;rep(i,0,n*2-1)rep(j,0,n*2-1)d.pb(Ppp(Pii(i,j),Pii(i,j)));
	solve(d1,d2,d);
	rep(i,0,n*n-1)printf("%d %d\n",res[i].fi,res[i].se);
	return 0;
}