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 <iostream>
using namespace std;
int n, mp[3][1003][1003];
void solve(int d, int w){
	int tmp[602][602];
	if (d % 4 == 0){
		solve(d / 4, w);
		for (int i = 0; i < 2 * n; ++ i)
			for (int j = 0; j < 2 * n; ++ j)
				tmp[i][j] = mp[w][i >> 1][j >> 1];
		for (int i = 0; i < 2 * n; ++ i)
			for (int j = 0; j < 2 * n; ++ j)
				mp[w][i][j] = tmp[i][j];
		return;
	}
	else{
		if (d % 2)
			for (int i = 0; i < 2 * n; ++ i)
				for (int j = 0; j < 2 * n; ++ j)
					mp[w][i][j] = (i + j + 1) & 1;
		else
			for (int i = 0; i < 2 * n; ++ i)
				for (int j = 0; j < 2 * n; ++ j)
					mp[w][i][j] = (i + 1) & 1;
	}
}
int main(){
	int d1, d2;
	cin >> n >> d1 >> d2;
	solve(d1, 1);
	solve(d2, 2);
	int cnt = 0;
	for (int i = 0; i < 2 * n; ++ i)
		for (int j = 0; j < 2 * n; ++ j)
			if (mp[1][i][j] && mp[2][i][j]){
				cout << i << " " << j << endl;
				if (++ cnt == n * n)
					return 0;
			}
}
