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;

typedef pair<int,int> pi;

const int maxn = 310;

int n, D1, D2, c1 = 0, c2 = 0;
vector<pi> vec[4];
int __s[4];

int getbit(int x, int k) {
	return (x >> k) & 1;
}

int main() {
	scanf("%d%d%d", &n, &D1, &D2);
	while (D1 % 2 == 0) {
		D1 /= 2;
		++ c1;
	}
	while (D2 % 2 == 0) {
		D2 /= 2;
		++ c2;
	}
	if (c1 == c2) {
		for (int x = 0; x < 2 * n; x++) {
			for (int y = 0; y < 2 * n; y++) {
				int v = 0;
				for (int i = 0; i < 10; i++) {
					v ^= (getbit(x, i) << (2 * i));
					v ^= (getbit(x, i) << (2 * i + 1));
					v ^= (getbit(y, i) << (2 * i));
				}
				vec[getbit(v, c1)].push_back(pi(x, y));
			}
		}
	} else {
		if (c1 > c2) {
			swap(c1, c2);
		}
		for (int x = 0; x < 2 * n; x++) {
			for (int y = 0; y < 2 * n; y++) {
				int v = 0;
				for (int i = 0; i < 10; i++) {
					v ^= (getbit(x, i) << (2 * i));
					v ^= (getbit(x, i) << (2 * i + 1));
					v ^= (getbit(y, i) << (2 * i));
				}
				vec[getbit(v, c1)*2+getbit(v, c2)].push_back(pi(x, y));
			}
		}
	}
	for (int i = 0; i < 4; i++) {
		__s[i] = int (vec[i].size());
	}
	int r = int (max_element(__s, __s + 4) - __s);
	for (int i = 0; i < n * n; i++) {
		printf("%d %d\n", vec[r][i].first, vec[r][i].second);
	}
	return 0;
}