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=1050;
int d[5],p[3][N][N],col[5][5],n;
int main(){
	cin>>n>>d[0]>>d[1];
	for(int i=0;i<2;i++){
		int a=0,b=0;
		for(int l=0;l<=1000;l++){
			for(int r=0;r<=1000;r++){
				if(l*l+r*r==d[i]){
					a=l,b=r;
					break;
				}
			}
		}
		//cerr<<a<<" "<<b<<endl;
		if(!a)continue;
		int cnt=1,op=0;
		while(a%2==0&&b%2==0){
			a/=2,b/=2,cnt*=2;
		}
		if((a&1)&&(b&1))op=1;
		else op=2;
		//cerr<<op<<" "<<cnt<<endl;
		for(int l=0;l<2*n;l++){
			for(int r=0;r<2*n;r++){
				int x=l/cnt,y=r/cnt;
				if(op==2){
					if((x+y)%2==0)p[i][l][r]=1;
					else p[i][l][r]=0;
				}
				else{
					if(x%2==0)p[i][l][r]=1;
					else p[i][l][r]=0;
				}
			}
		}
	}
	for(int i=0;i<2*n;i++){
		for(int j=0;j<2*n;j++){
			col[p[0][i][j]][p[1][i][j]]++;
		}
	}
	for(int i=0;i<2;i++){
		for(int j=0;j<2;j++){
			if(col[i][j]>=n*n){
				int cnt=n*n;
				for(int l=0;l<2*n;l++){
					for(int r=0;r<2*n;r++){
						if(p[0][l][r]==i&&p[1][l][r]==j){
							printf("%d %d\n",l,r);
							--cnt;
							if(cnt==0)exit(0);
						}
					}
				}
			}
		}
	}
}
