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 maxn = 305;
int vis[maxn<<1][maxn<<1][2], color[maxn<<1][maxn<<1][2];
int N,D1,D2;
int tot,color_cnt;
typedef pair<int,int> p;
vector<p> v1,v2;

void dfs(int x,int y,int a,int b){
	if(x < 0 || x >= 2 * N || y < 0 || y >= 2 * N || vis[x][y][a]) return;
	vis[x][y][a] = 1; color[x][y][a] = b;
	for(int i = 0; i < ((a == 0) ? v1.size() : v2.size()); i++){
		int dx = ((a == 0) ? v1[i].first : v2[i].first);
		int dy = ((a == 0) ? v1[i].second : v2[i].second);
		dfs(x + dx,y + dy, a, 3 - b);
		dfs(x - dx,y + dy, a, 3 - b);
		dfs(x + dx,y - dy, a, 3 - b);
		dfs(x - dx,y - dy, a, 3 - b);
	} 
}

int main(){
	scanf("%d%d%d",&N,&D1,&D2);
	// Decompose
	for(int i = 0; i <= D1; i++){
		int y = (int) sqrt(D1 - i * i);
		if(y * y + i * i == D1) v1.push_back(p(i,y));
	}
	for(int i = 0; i <= D2; i++){
		int y = (int) sqrt(D2 - i * i);
		if(y * y + i * i == D2) v2.push_back(p(i,y));
	}
	for(int i = 0; i < N * 2; i++){
		for(int j = 0; j < N * 2; j++){
			if(!vis[i][j][0]) dfs(i,j,0,1);
			if(!vis[i][j][1]) dfs(i,j,1,1);
		}
	}
	int cnt = 0;
	for(int i = 0; i < N * 2; i++){
		for(int j = 0; j < N * 2; j++){
			if(cnt == N * N) return 0;
			if(color[i][j][0] == 1 && color[i][j][1] == 1){
				printf("%d %d\n",i,j);
				cnt++; 
			}
		}
	}
	return 0;
}
