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 REP(i,a,b) for(int i=(a),_ed=(b);i<=_ed;++i)
#define DREP(i,a,b) for(int i=(a),_ed=(b);i>=_ed;--i)
#define mp(x,y) make_pair((x),(y))
#define sz(x) (int)(x).size()
#define pb push_back
typedef long long ll;
typedef pair<int,int> pii;
inline int read(){
    register int x=0,f=1;register char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=0;ch=getchar();}
    while(isdigit(ch)){x=x*10+(ch^'0');ch=getchar();}
    return f?x:-x;
}

int n,col[605][605];

void paint(int d){
	int cnt=0;
	while(d%4==0)d/=4,++cnt;
	if(d&1){
		REP(i,0,n+n-1)REP(j,0,n+n-1)
			col[i][j]|=((i>>cnt)+(j>>cnt))&1;
	}
	else{
		REP(i,0,n+n-1)REP(j,0,n+n-1)
			col[i][j]|=(i>>cnt)&1;
	}
}

int main(){
	// freopen("in.in","r",stdin);
	n=read();
	paint(read()),paint(read());
	int cnt=0;
	REP(i,0,n+n-1)REP(j,0,n+n-1){
		if(!col[i][j])printf("%d %d\n",i,j),++cnt;
		if(cnt==n*n)return 0;
	}
	return 0;
}
