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<iostream>
#include<cstdio>
#include<cmath>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=1010;
int n,D1,D2,K,m,cnt,c[2][N][N],s[4];
void work(int L,int C[N][N]){
	L=log2(L&-L);
	K=!(L&1);L>>=1;
	FOR(i,0,m)FOR(j,0,m) C[i][j]=i+K*j>>L&1;
}
int main(){
	//freopen("1.in","r",stdin);
	scanf("%d%d%d",&n,&D1,&D2);
	m=2*n-1;
	work(D1,c[0]);work(D2,c[1]);
	FOR(i,0,m)FOR(j,0,m) s[c[0][i][j]|c[1][i][j]<<1]++;
	FOR(k,0,3)if(s[k]>=n*n){
		FOR(i,0,m)FOR(j,0,m)if((c[0][i][j]|c[1][i][j]<<1)==k && cnt<n*n)
			cout<<i<<' '<<j<<'\n',++cnt;
		return 0;
	} 
}