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 <cstdio>
#include <vector>
#include <iostream>
using namespace std;

const int MAXN = 600;
const int MAXM = MAXN*MAXN;

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

bool clr[20][MAXM + 5];
int x[MAXM + 5], y[MAXM + 5], id[MAXN + 5][MAXN + 5], cnt;
int main() {
	int N, D1, D2, p1 = 0, p2 = 0; scanf("%d%d%d", &N, &D1, &D2);
	while( D1 % 2 == 0 ) D1 /= 2, p1++;
	while( D2 % 2 == 0 ) D2 /= 2, p2++;
	for(int i=0;i<2*N;i++)
		for(int j=0;j<2*N;j++)
			id[i][j] = (++cnt), x[cnt] = i, y[cnt] = j;
	for(int o=0;o<20;o++)
		for(int i=1;i<=cnt;i++) {
			clr[o][i] = ((x[i] + y[i]) % 2 != 0);
			int tx = (x[i] + y[i] + clr[o][i]) / 2, ty = (x[i] - y[i] + clr[o][i]) / 2;
			x[i] = tx, y[i] = ty;
		}
	for(int i=0;i<2*N;i++)
		for(int j=0;j<2*N;j++) {
			int x = id[i][j];
			ans[clr[p1][x] | (clr[p2][x] << 1)].push_back(make_pair(i, j));
		}
	int res = 0;
	for(int i=0;i<4;i++)
		if( ans[i].size() > ans[res].size() )
			res = i;
	for(int i=0;i<N*N;i++)
		printf("%d %d\n", ans[res][i].first, ans[res][i].second);
	
}