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;
const ll INF = 1LL << 60;
const ll MOD = 1e9 + 7;

#define rep(i,n) for (int i = 0; i < (n); i++)
#define ALL(v) v.begin(), v.end()
typedef pair<int, int> P;
typedef pair<P, int> PI;

void color(vector<PI> &points, vector<int> &c, int D) {
	if (D % 2 == 1) {
		for (PI p : points) {
			c[p.second] = (p.first.first + p.first.second) % 2;
		}
	}
	else if (D % 4 == 2) {
		for (PI p : points) {
			c[p.second] = p.first.first % 2;
		}
	}
	else {
		vector<PI> v[2][2];
		for (PI p : points) {
			v[p.first.first % 2][p.first.second % 2].emplace_back(P(p.first.first / 2, p.first.second / 2), p.second);
		}
		for (int i = 0; i < 2; i++) {
			for (int j = 0; j < 2; j++) color(v[i][j], c, D / 4);
		}
	}
}
int main() {
	int N, D1, D2;
	cin >> N >> D1 >> D2;
	vector<PI> points;
	vector<int> c1(4 * N*N), c2(4 * N*N);
	for (int i = 0; i < 2 * N; i++) {
		for (int j = 0; j < 2 * N; j++) {
			points.emplace_back(P(i, j), i*(2 * N) + j);
		}
	}
	color(points, c1, D1);
	color(points, c2, D2);
	vector<PI> group[4];
	for (PI p : points) {
		group[c1[p.second] * 2 + c2[p.second]].push_back(p);
	}
	for (int i = 0; i < 4; i++) {
		if (group[i].size() >= N * N) {
			for (int j = 0; j < N*N; j++) {
				cout << group[i][j].first.first << " " << group[i][j].first.second << endl;
			}
			return 0;
		}
	}
}
