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 INF int(1e9)
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=305;
const int maxm=maxn*maxn*4;
int n,len;
pii val[maxm];
vector<int>edge[maxm];
int col[maxm],cnt[4];
bool vis[maxm];

void dfs(int x,int tp,int c){
	vis[x]=true;
	col[x]|=((1<<tp)*c);
	rep(i,int(edge[x].size())){
		int y=edge[x][i];
		if(!vis[y])dfs(y,tp,c^1);
	}
}

int main(){
	scanf("%d",&n);
	rep(tp,2){
		int d;
		scanf("%d",&d);
		rep(i,4*n*n){
			edge[i].clear();
			vis[i]=false;
		}
		len=0;
		rep(i,2*n+1){
			if(i*i>d)break;
			int j=sqrt(d-i*i);
			if(j*j+i*i==d&&j<=2*n){
				val[len++]=mp(i,j);
				if(i)val[len++]=mp(-i,j);
				if(j)val[len++]=mp(i,-j);
				if(i&&j)val[len++]=mp(-i,-j);
			}
		}
		rep(i,2*n)rep(j,2*n){
			rep(k,len){
				int ni=i+val[k].fst,nj=j+val[k].snd;
				if(ni>=0&&ni<2*n&&nj>=0&&nj<2*n)edge[i*2*n+j].pb(ni*2*n+nj);
			}
		}
		rep(i,4*n*n)if(!vis[i])dfs(i,tp,0); 
	}
	rep(i,4*n*n)cnt[col[i]]++;
	int mxid=-1;
	rep(i,4)if(mxid<0||cnt[i]>cnt[mxid])mxid=i;
	int tot=0;
	rep(i,2*n)rep(j,2*n)if(col[i*2*n+j]==mxid){
		printf("%d %d\n",i,j);
		tot++;
		if(tot==n*n)return 0;
	}
	return 0;
}