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;
		vector<vector<int>> A(2*N, vector<int>(2*N));
		{
			int n = 2;
			int x = 1;
			for (;;) {
				if (D1 % n) {
					for (int i = 0; i < 2*N; ++ i) for (int j = 0; j < 2*N; ++ j) A[i][j] |= (i/x + j/x + 1) % 2;
					break;
				}
				n *= 2;
				if (D1 % n) {
					for (int i = 0; i < 2*N; ++ i) for (int j = 0; j < 2*N; ++ j) A[i][j] |= (i/x + 1) % 2;
					break;
				}
				n *= 2;
				x *= 2;
			}
		}
		{
			int n = 2;
			int x = 1;
			for (;;) {
				if (D2 % n) {
					for (int i = 0; i < 2*N; ++ i) for (int j = 0; j < 2*N; ++ j) A[i][j] |= (i/x + j/x + 1) % 2 * 2;
					break;
				}
				n *= 2;
				if (D2 % n) {
					for (int i = 0; i < 2*N; ++ i) for (int j = 0; j < 2*N; ++ j) A[i][j] |= (j/x + 1) % 2 * 2;
					break;
				}
				n *= 2;
				x *= 2;
			}
		}
		int cnt = 0;
		for (int i = 0; i < 2*N; ++ i) for (int j = 0; j < 2*N; ++ j) if (A[i][j] == 3) {
			cout << i << " " << j << endl;
			if (++cnt == N*N) return 0;
		}
	}
