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>

const int N = 600 + 7;

int n;
int f[N][N], g[4], h;

int main() {
//	freopen("code.in", "r", stdin);
//	freopen("code.out", "w", stdout);

	scanf("%d", &n), n <<= 1;
	for (int t = 1, k, d; t <= 2; ++t) {
		scanf("%d", &d),
		k = 1 << __builtin_ctz(d) / 2;
		d = d / k / k % 4 == 1;
		for (int i = 0; i < n; ++i)
			for (int j = 0; j < n; ++j)
				f[i][j] |= (i / k + j / k * d) & 1 ? t : 0;
	}
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < n; ++j)
			++g[f[i][j]];
	for (int i = 1; i <= 3; ++i)
		h = g[i] > g[h] ? i : h;
	for (int i = 0, k = n * n / 4; i < n; ++i)
		for (int j = 0; j < n; ++j)
			if (f[i][j] == h && k) --k, printf("%d %d\n", i, j);

	return 0;
}
