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, A[2];
bool ok[1000][1000];

int main() {
	cin >> N >> A[0] >> A[1];
	for (int i = 0; i < 1000; ++i) for (int j = 0; j < 1000; ++j) ok[i][j] = true;

	// 2 で何回割れるか
	int num[2] = { 0 };
	for (int iter = 0; iter < 2; ++iter) {
		num[iter] = 0;
		while (A[iter] % 2 == 0) {
			++num[iter];
			A[iter] /= 2;
		}
	}

	// 盤面をふるいにかける
	for (int iter = 0; iter < 2; ++iter) {
		int p = num[iter];
		if (p & 1) {
			int two = 1 << (p / 2);
			for (int x = 0; x < N * 2; ++x) {
				for (int y = 0; y < N * 2; ++y) {
					if (y % (two * 2) >= two) ok[x][y] = false;
				}
			}
		}
		else {
			int two = 1 << (p / 2);
			for (int x = 0; x < N * 2; ++x) {
				for (int y = 0; y < N * 2; ++y) {
					int px = (x % (two * 2) >= two ? 1 : 0);
					int py = (y % (two * 2) >= two ? 1 : 0);
					if (px != py) ok[x][y] = false;
				}
			}
		}
	}

	// N^2 個になるまで出力
	int con = 0;
	for (int x = 0; x < N * 2; ++x) {
		for (int y = 0; y < N * 2; ++y) {
			if (con == N*N) break;
			if (ok[x][y]) {
				cout << x << " " << y << endl;
				++con;
			}
		}
	}
}