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;
typedef long long LL;
int ok[1000][1000];

vector<vector<int> > color(int n, int d){
	vector<vector<int> > g(2*n);
	for(int j = 0; j < 2*n; j++){
		g[j].resize(2*n, 0);
	}
	int b = 0;
	while(d % (1<<b) == 0) b++;
	if(b % 2 == 1){
		for(int i = 0; i < 2*n; i++){
			for(int j = 0; j < 2*n; j++){
				int c = (b-1)/2;
				int num = (i>>c) + (j>>c);
				g[i][j] = num % 2;
			}
		}
	} else if(b % 2 == 0){
		for(int i = 0; i < 2*n; i++){
			for(int j = 0; j < 2*n; j++){
				int c = (b-2)/2;
				int num = (i>>c);
				g[i][j] = num % 2;
			}
		}
	}
	return g;
}
int main(){
	cin.sync_with_stdio(0); cin.tie(0);
	int n;
	int d1, d2;
	cin >> n >> d1 >> d2;
	vector<vector<int> > c1 = color(n, d1);
	vector<vector<int> > c2 = color(n, d2);
	vector<vector<int> > c3 = c1;
	vector<pair<int,int> > g[4];
	for(int i = 0; i < 2*n; i++){
		for(int j = 0; j < 2*n; j++){
			g[c1[i][j] * 2 + c2[i][j]].push_back({i,j});
		}
	}
	for(int j = 0; j < 4; j++){
		if(g[j].size() >= n*n){
			for(int i = 0; i < n*n; i++){
				cout << g[j][i].first << " " << g[j][i].second << '\n';
			}
			return 0;
		}
	}
}
