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>
#define ctz __builtin_ctz

const int N = 648;

int n, D1, D2;
int a[N][N];

void coloring(int dist, int bit) {
	int i, j, L = ctz(dist) / 2; dist >>= 2 * L;
	switch (dist & 3) {
		case 0: throw "gg"; break;
		case 1: for (i = 0; i < n; ++i) for (j = 0; j < n; ++j) a[i][j] |= ((i ^ j) >> L & 1) << bit; break;
		case 2: for (i = 0; i < n; ++i) for (j = 0; j < n; ++j) a[i][j] |= (i >> L & 1) << bit; break;
	}
}

int main() {
	int i, j, c, d, col[4] = {0};
	scanf("%d%d%d", &n, &D1, &D2), d = n * n, n *= 2, coloring(D1, 0), coloring(D2, 1);
	for (i = 0; i < n; ++i)
		for (j = 0; j < n; ++j) ++col[ a[i][j] ];
	c = std::max_element(col, col + 4) - col;
	for (i = 0; i < n && d; ++i)
		for (j = 0; j < n && d; ++j)
			if (a[i][j] == c) --d, printf("%d %d\n", i, j);
	return 0;
}