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 rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define qwq(x) cout<<"# "<<#x<<" = "<<x<<endl;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define f(x,y) g[(x)][(y)+52]
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
 
template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=1005;
int f[N][N],n,d1,d2,cnt;

void doit(int d){
	int k=1;
	while(d%4==0)d>>=2,k<<=1;
	REP(i,n*2)REP(j,n*2)
		f[i][j]+=(d&1)&&(i/k+j/k)%2==0||!(d&1)&&i/k%2==0;
}

int main(){
	read(n),read(d1),read(d2);
	doit(d1),doit(d2);
	REP(i,n*2)REP(j,n*2)
		if(f[i][j]==2&&cnt<n*n)
			printf("%d %d\n",i,j),++cnt;
	return 0;
}