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;

int n, d1, d2;
vector<pair<int, int> > v[2], a[2];

int main(){
	scanf("%d%d%d", &n, &d1, &d2);
	int x = 0, y;
	while (d1%(1<<(2*(x+1)))==0) x++;
	x=(1<<x);
	y = d1 / (x * x) % 4;
	//x = max(x / 2, 1);
	for(int i = 0;i < n * 2;i++){
		for(int j = 0;j < n * 2;j++){
			int ii = i / x, jj = j / x;
			if(y == 1){
				v[(ii + jj) & 1].push_back({i, j});
			}else if(y == 2){
				v[ii & 1].push_back({i, j});
			}else{
				v[0].push_back({i, j});
			}
		}
	}
	if(v[0].size() < v[1].size()) swap(v[0], v[1]);
	x = 0;
	while (d2%(1<<(2*(x+1)))==0) x++;
	x=(1<<x);
	y = d2 / (x * x) % 4;
	//x = max(x / 2, 1);
	for(int k = 0;k < v[0].size();k++){
		int i = v[0][k].first, j = v[0][k].second;
		int ii = i / x, jj = j / x;
		if(y == 1){
			a[(ii + jj) & 1].push_back({i, j});
		}else if(y == 2){
			a[ii & 1].push_back({i, j});
		}else{
			a[0].push_back({i, j});
		}
	}
	if(a[0].size() < a[1].size()) swap(a[0], a[1]);
	for(int i = 1;i <= n * n;i++) printf("%d %d\n", a[0].back().first, a[0].back().second), a[0].pop_back();
}