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 pb  push_back
#define mp  make_pair
#define v1  first
#define v2  second
using namespace std;

typedef pair <int,int>  PII;
const int N=605;
int n,d1,d2,vis[N][N];
vector <PII> V;

void dfs(int x,int y)
{
	for (auto d:V)
		if ((0<=x+d.v1)&&(x+d.v1<2*n)&&(0<=y+d.v2)&&(y+d.v2<2*n))
			if (!vis[x+d.v1][y+d.v2])
				{
					vis[x+d.v1][y+d.v2]=vis[x][y]^1;
					dfs(x+d.v1,y+d.v2);
				}
}

void solve(int d)
{
	V.clear();
	for (int i=-2*n+1; i<2*n; i++)
		for (int j=-2*n+1; j<2*n; j++)
			if (i*i+j*j==d)
				V.pb(mp(i,j));
	for (int i=0; i<2*n; i++)
		for (int j=0; j<2*n; j++)
			if (!vis[i][j])
				vis[i][j]=2,dfs(i,j);
	int cnt2=0,cnt3=0;
	for (int i=0; i<2*n; i++)
		for (int j=0; j<2*n; j++)
			{
				cnt2+=(vis[i][j]==2);
				cnt3+=(vis[i][j]==3);
			}
	if (cnt2<cnt3)
		{
			for (int i=0; i<2*n; i++)
				for (int j=0; j<2*n; j++)
					if (vis[i][j]>1)
						vis[i][j]^=1;
		}
	for (int i=0; i<2*n; i++)
		for (int j=0; j<2*n; j++)
			if (vis[i][j]>1)
				vis[i][j]-=2;
}

void work()
{
	scanf("%d %d %d",&n,&d1,&d2);
	solve(d1),solve(d2);
	for (int i=0,k=n*n; i<2*n; i++)
		for (int j=0; j<2*n; j++)
			if ((!vis[i][j])&&(k>0))
				printf("%d %d\n",i,j),k--;
}

int main()
{
	work();
	return 0;
}
