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;
 
#define pb push_back
#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)
 
#define N (600 + 5)
 
struct node{int a, b;};
vector<node> ans[4];
 
bool vis[2][N][N];
int n, nn, t1, t2, t3, t4, d[2], col[2][N][N];
 
void Dfs(int x, int y, int o, int c){
	if(x > nn || y > nn || x < 0 || y < 0 || col[o][x][y]) return;
	vis[o][x][y] = true; col[o][x][y] = 3 - c;
 
	int a = -1;
	while(true){
		++a;
		if(a * a > d[o]) return;
 
		int b = sqrt(d[o] - a * a);
		if(a * a + b * b != d[o]) continue;
		
		Dfs(x + a, y + b, o, 3 - c); Dfs(x + a, y - b, o, 3 - c);
		Dfs(x - a, y - b, o, 3 - c); Dfs(x - a, y + b, o, 3 - c);
	}
}
 
void Print(int o){
	For(i, 0, n * n - 1) printf("%d %d\n", ans[o][i].a, ans[o][i].b);	
}
 
int main(){
	scanf("%d%d%d", &n, &d[0], &d[1]);
 
	nn = (n << 1) - 1;
	For(i, 0, nn) For(j, 0, nn) if(!vis[0][i][j]) Dfs(i, j, 0, 2);
	For(i, 0, nn) For(j, 0, nn) if(!vis[1][i][j]) Dfs(i, j, 1, 2);
	
	For(i, 0, nn) For(j, 0, nn){
		node o = (node){i, j};
 
		if(col[0][i][j] == 1){
			if(col[1][i][j] == 1) ans[0].pb(o); else ans[1].pb(o);
		}else{
			if(col[1][i][j] == 2) ans[2].pb(o); else ans[3].pb(o);
		}
	}
 
	if((int)ans[0].size() >= n * n) Print(0);
	else if((int)ans[1].size() >= n * n) Print(1);
	else if((int)ans[2].size() >= n * n) Print(2);
	else Print(3);
	
	return 0;
}