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 x[666][666];
int main()
{
	int n, d[2];
	cin >> n >> d[0] >> d[1];
	for (int i = 0; i < 2; i++) {
		int e = 0;
		for (; !(d[i] & 1); e++)
			d[i] >>= 1;
		for (int j = 0; j < n * 2; j++) {
			for (int k = 0; k < n * 2; k++)
				if ((j + k - e % 2 * k) >> e / 2 & 1)
					x[j][k] |= i + 1;
		}
	}
	int c[4] = {};
	for (int j = 0; j < n * 2; j++)
		for (int k = 0; k < n * 2; k++)
			c[x[j][k]]++;
	int m = 0;
	for (int i = 0; i < 4; i++)
		if (c[i] > c[m]) m = i;
	int i = 0;
	for (int j = 0; j < n * 2; j++) {
		for (int k = 0; k < n * 2; k++) {
			if (x[j][k] != m)
				continue;
			if (i++ >= n * n)
				goto pnz;
			cout << j << ' ' << k << '\n';
		}
	}
pnz:
	cout << flush;
	return 0;
}
