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 ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=605;
bool a[20][N][N];
bool ans[N][N];
int n,D1,D2;
void work(int n,int D,int d){
	if (D%4){
		if (D%2){
			For(i,0,n-1) For(j,0,n-1)
				a[d][i][j]=(i+j+1)&1;
		}
		else{
			For(i,0,n-1) For(j,0,n-1)
				a[d][i][j]=(i+1)&1; 
		}
		return;
	}
	work(n,D/4,d+1);
	For(i,0,n-1) For(j,0,n-1)
		a[d][i][j]=a[d+1][i/2][j/2];
}
int main(){
	scanf("%d%d%d",&n,&D1,&D2);
	work(2*n,D1,0);
	For(i,0,2*n-1) For(j,0,2*n-1)
		ans[i][j]=a[0][i][j];
	work(2*n,D2,0);
	For(i,0,2*n-1) For(j,0,2*n-1)
		ans[i][j]&=a[0][i][j];
	int rest=n*n;
	For(i,0,2*n-1) For(j,0,2*n-1)
		if (rest&&ans[i][j])
			--rest,printf("%d %d\n",i,j);
}