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;
typedef pair<int,int>pii;
const int N=605;
int n,D1,D2,i,j,k,l,c[N][N],ct,ctt[2];
bool b[N][N];
vector<pii>v[2],dt;
int ZZ=0,Z2=0;
void dfs(int x,int y,int o){
	if(x<0 || x>=n*2 || y<0 || y>=n*2 || c[x][y]!=-1)return;
	v[o].push_back(pii(x,y));c[x][y]=o;ctt[o]+=b[x][y];
	for(pii z:dt){
		int i=z.first,j=z.second;
		dfs(x+i,y+j,!o);
		dfs(x-j,y+i,!o);
		dfs(x-i,y-j,!o);
		dfs(x+j,y-i,!o);
	}
}
int main(){
	scanf("%d%d%d",&n,&D1,&D2);
	memset(b,1,sizeof b);
	for(i=1;i<N;++i)for(j=0;j<N;++j)if(i*i+j*j==D1)dt.push_back(pii(i,j));
	memset(c,-1,sizeof c);
	for(k=0;k<n*2;++k)for(l=0;l<n*2;++l)if(c[k][l]==-1){
		ctt[0]=ctt[1]=0;
		v[0].clear(),v[1].clear(),dfs(k,l,0);
		if(ctt[0]>ctt[1])swap(v[0],v[1]);
		for(pii u:v[0])b[u.first][u.second]=0;
	}
	dt.clear();
	for(i=1;i<N;++i)for(j=0;j<N;++j)if(i*i+j*j==D2)dt.push_back(pii(i,j));
	memset(c,-1,sizeof c);
	for(k=0;k<n*2;++k)for(l=0;l<n*2;++l)if(c[k][l]==-1){
		ctt[0]=ctt[1]=0;
		v[0].clear(),v[1].clear(),dfs(k,l,0);
		if(ctt[0]>ctt[1])swap(v[0],v[1]);
		for(pii u:v[0])b[u.first][u.second]=0;
	}
	k=n*n;
	for(i=0;i<n*2;++i)for(j=0;j<n*2;++j)if(b[i][j] && k)--k,printf("%d %d\n",i,j);
assert(!k);
	return 0;
}