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;
#define ll long long
#define ui long long
#define ull unsigned long long
#define db long double
#define pii pair<ll,ll>
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define vi vector<int>
#define vii vector<vi>
#define lb lower_bound
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,b,a) for(int i=(b);i>=(a);--i)
#define rep0(i,a,b) for(int i=(a);i<(b);++i)
#define fore(i,a) for(int i=0;i<a.size();++i)
#define gc() getchar()
#define ls x<<1,l,m
#define rs x<<1|1,m+1,r
inline int rd()
{
	int x=0;char c=gc();while(!isdigit(c))c=gc();
	while(isdigit(c))x=x*10+c-48,c=gc();return x;
}
const int N=605;
int n,m,d1,d2,c[N][N];
void sol(int d)
{
	int t=0;while(!(d&3))d>>=2,t++;
	if(d&1){rep0(i,0,m)rep0(j,0,m)if(((i>>t)+(j>>t))&1)c[i][j]=1;}
	else{rep0(i,0,m)rep0(j,0,m)if(i>>t&1)c[i][j]=1;}
}
int main()
{
	n=rd();d1=rd();d2=rd();m=n*2;
	sol(d1);sol(d2);int nn=n*n;
	rep0(i,0,m)rep0(j,0,m)if(!c[i][j])
	{
		printf("%d %d\n",i,j);nn--;
		if(!nn)return 0;
	}
	return 0;
}