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 MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
int d1,d2;
vector<P> ng[2];
int col[601][601];
bool used[601][601];

void dfs(int y,int x,int t,int c){
	used[y][x]=true;
	if(c==1)col[y][x]+=1<<t;
	for(int i=0;i<ng[t].size();i++){
		int nx=ng[t][i].first;
		int ny=ng[t][i].second;
		for(int j=0;j<4;j++){
			if(nx+x>=0 && nx+x<2*n && ny+y>=0 && ny+y<2*n){
				if(!used[ny+y][nx+x]){
					dfs(ny+y,nx+x,t,1-c);
				}
			}
			if(j==0)nx=-nx;
			if(j==1)ny=-ny;
			if(j==2)nx=-nx;
		}
	}
}

int main(void){
	scanf("%d%d%d",&n,&d1,&d2);
	if(d1>d2){
		swap(d1,d2);
	}
	for(int i=0;i<=n*2;i++){
		for(int j=0;j<=n*2;j++){
			int x=i*i+j*j;
			if(x==d1){
				ng[0].push_back(P(i,j));
			}
			if(x==d2){
				ng[1].push_back(P(i,j));
			}
		}
	}
	for(int k=0;k<2;k++){
		memset(used,false,sizeof(used));
		for(int i=0;i<n*2;i++){
			for(int j=0;j<n*2;j++){
				if(!used[i][j]){
					dfs(i,j,k,0);
				}
			}
		}
	}
	for(int i=0;i<4;i++){
		int cnt=0;
		for(int j=0;j<2*n;j++){
			for(int k=0;k<2*n;k++){
				if(i==col[j][k])cnt++;
			}
		}
		if(cnt>=n*n){
			int cl=0;
			for(int j=0;j<2*n;j++){
				for(int k=0;k<2*n;k++){
					if(cl==n*n)break;
					if(i==col[j][k]){
						cl++;
						printf("%d %d\n",k,j);
					}	
				}
			}
			break;
		}
	}
	return 0;
}