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 d1, d2, n;
vector<pair<int, int>> s1, s2;

int c1[600][600], c2[600][600];

template<class T, class U>
void boji(T c, U& s, int x, int y) {
	for (auto p : s)
		for (int dx=-1; dx<=1; dx+=2)
			for (int dy=-1; dy<=1; dy+=2) {
				int xx = x + dx*p.first;
				int yy = y + dy*p.second;

				if (xx < 0 || xx >= 2*n || yy < 0 || yy >= 2*n)
					continue;

				if (!c[xx][yy]) {
					c[xx][yy] = 3 - c[x][y];
					boji(c, s, xx, yy);
				}
			}
}

vector<pair<int, int>> sx[4];

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cerr.tie(nullptr);

	cin >> n >> d1 >> d2;
	for (int i=0; i<=2*n; i++) {
		for (int j=0; j<=2*n; j++) {
			if (i*i + j*j == d1)
				s1.push_back({i, j});
			if (i*i + j*j == d2)
				s2.push_back({i, j});
		}
	}

	for (int i=0; i<2*n; i++)
		for (int j=0; j<2*n; j++)
			if (!c1[i][j]) {
				c1[i][j] = 1;
				boji(c1, s1, i, j);
			}

	for (int i=0; i<2*n; i++)
		for (int j=0; j<2*n; j++)
			if (!c2[i][j]) {
				c2[i][j] = 1;
				boji(c2, s2, i, j);
			}

	for (int i=0; i<2*n; i++)
		for (int j=0; j<2*n; j++) {
			int x = c1[i][j] + 2*c2[i][j] - 3;
			sx[x].push_back({i, j});
		}

	for (int i=0; i<4; i++) {
		if (sx[i].size() >= n*n) {
			for (int j=0; j<n*n; j++) 
				cout << sx[i][j].first << ' ' << sx[i][j].second << '\n';
			return 0;
		}
	}
}