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;

const int N = 310;
int tot;
vector<pair<int, int> > ans[2][2];

void dfs(int c[N * 2][N * 2], int x, int y, int n, int d, int col) {
	if (x > n || y > n || x < 0 || y < 0 || c[x][y] != -1) return;
	c[x][y] = col;
	for (int i = 0; i * i <= d && i <= n; i++) {
		int j = sqrt(d + 0.0 - i * i);
		while (j * j <= d - i * i) j++;
		j--;
		if (i * i + j * j != d) continue;
		dfs(c, x + i, y + j, n, d, col ^ 1);
		dfs(c, x + i, y - j, n, d, col ^ 1);
		dfs(c, x - i, y + j, n, d, col ^ 1);
		dfs(c, x - i, y - j, n, d, col ^ 1);
	}
}

int c[2][N * 2][N * 2];
void solve(int n, int d, int t) {
	for (int i = 0; i < n * 2; i++)
		for (int j = 0; j < n * 2; j++)
			if (c[t][i][j] == -1) dfs(c[t], i, j, n * 2 - 1, d, 0);
}

int main()
{
	int n, d1, d2;
	scanf("%d%d%d", &n, &d1, &d2);
    memset(c, -1, sizeof(c));
	solve(n, d1, 0);
	solve(n, d2, 1);
	for (int i = 0; i < 2 * n; i++)
		for (int j = 0; j < 2 * n; j++) {
			ans[c[0][i][j]][c[1][i][j]].push_back(make_pair(i, j));
		}
	tot = n * n;
	int maxn = 0, x = 0, y = 0;
    for (int i = 0; i < 2; i++)
		for (int j = 0; j < 2; j++)
            if (ans[i][j].size() > maxn) maxn = ans[i][j].size(), x = i, y = j;
	for (int i = 0; i < ans[x][y].size() && i < tot; i++)
		printf("%d %d\n", ans[x][y][i].first, ans[x][y][i].second);
	return 0;
}
