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 ll long long
#define pb push_back
#define mr make_pair
using namespace std;
template <typename T> void read(T &x){
	x=0;char ch=getchar();int fh=1;
	while (ch<'0'||ch>'9'){if (ch=='-')fh=-1;ch=getchar();}
	while (ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
	x*=fh;
}
vector<pair<int,int> > a[2][2];
int n,d1,b[610][610];
void fe(int x,int nu){
	int s1=0,s2=x;
	while (s2%2==0){s1++;s2/=2;}
	if (s1%2){
		s1/=2;
		for (int i=0;i<2*n;i++)
			for (int j=0;j<2*n;j++)
				a[nu][(i>>s1)%2].pb(mr(i,j));
	}else{
		s1/=2;
		for (int i=0;i<2*n;i++)
			for (int j=0;j<2*n;j++)
				a[nu][((i>>s1)+(j>>s1))%2].pb(mr(i,j));
	}
	if (a[nu][0].size()<a[nu][1].size())swap(a[nu][0],a[nu][1]);
}
int main(){
	read(n);
	read(d1);fe(d1,0);
	read(d1);fe(d1,1);
	for (auto x:a[0][0])b[x.first][x.second]++;
	for (auto x:a[1][0])b[x.first][x.second]++;
	for (int i=0,k=0;i<2*n;i++)
		for (int j=0;j<2*n;j++)if (b[i][j]==2&&k<n*n){cout<<i<<' '<<j<<'\n';k++;}
	return 0;
}