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 main() {
	int n, d1, d2;
	cin >> n >> d1 >> d2;
	int cnt1, cnt2;
	cnt1 = cnt2 = 0;
	while (d1 % 4 == 0) d1 /= 4, cnt1++;
	while (d2 % 4 == 0) d2 /= 4, cnt2++;
	int cnt = 0;
	for (int i = 0; i < n * 2; ++i) {
		for (int j = 0; j < n * 2; ++j) {
			if (cnt == n * n) continue;
			int x, y;
			x = i / (1 << cnt1);
			y = j / (1 << cnt1);
			if (d1 % 4 == 1 && (x + y) % 2 == 1) continue;
			if (d1 % 4 == 2 && x % 2 == 1) continue;
			x = i / (1 << cnt2);
			y = j / (1 << cnt2);
			if (d2 % 4 == 1 && (x + y) % 2 == 1) continue;
			if (d2 % 4 == 2 && x % 2 == 1) continue;
			cout << i << ' ' << j << '\n';
			++cnt;
		}
	}
}