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<cstring>
#include<algorithm>
#define clr(x) memset(x,0,sizeof(x))
#define MN 603
using namespace std;
inline int in(){
    int x=0;bool f=0;char c;
    for (;(c=getchar())<'0'||c>'9';f=c=='-');
    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');
    return f?-x:x;
}
const int dx[4]={1,1,-1,-1};
const int dy[4]={1,-1,1,-1};
int col[MN][MN],X[MN],Y[MN],num[4];
int n,m,d1,d2,tot,cnt,res;
bool vis[MN][MN];
inline bool check(int x,int y){
	return ((x>=0)&&(y>=0)&&(x<m)&&(y<m)&&(!vis[x][y]));
}
inline void dfs(int x,int y,bool c){
	vis[x][y]=1;col[x][y]=(col[x][y]<<1)+c;
	for (int i=0;i<cnt;++i)
	for (int j=0;j<4;++j)
	if (check(x+dx[j]*X[i],y+dy[j]*Y[i])) 
	dfs(x+dx[j]*X[i],y+dy[j]*Y[i],c^1);
	
}
inline void sol(int d){
	clr(X);clr(Y);clr(vis);cnt=0;
	for (int i=0;i<m;++i){
		if(i*i>d) break;
		for (int j=0;j<m;++j) 
		if (i*i+j*j==d) X[cnt]=i,Y[cnt]=j,++cnt;
	}
	for (int i=0;i<m;++i)
	for (int j=0;j<m;++j) if (!vis[i][j]) dfs(i,j,0);
}
int main()
{
	n=in();d1=in();d2=in();m=(n<<1);
	sol(d1);sol(d2);tot=0;
	for (int i=0;i<m;++i)
	for (int j=0;j<m;++j) ++num[col[i][j]];
	for (int i=0;i<4;++i) if (num[i]>=n*n) res=i;
	for (int i=0;i<m;++i)
	for (int j=0;j<m;++j){
		if (res==col[i][j]) ++tot,printf("%d %d\n",i,j);
		if (tot==n*n) return 0;
	}return 0;
}