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>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
using namespace std;
#define N 605
bool vis[N][N][2];
int n,cnt;
int main(){
	scanf("%d",&n);
	rep (op,0,1){
		int d,len=0;scanf("%d",&d);
		while (d%4==0) d/=4,len++;
		if (d%4==2){
			rep (i,0,2*n-1) rep (j,0,2*n-1)
				if (i>>len&1) vis[i][j][op]=1;
		} else{
			rep (i,0,2*n-1) rep (j,0,2*n-1)
				if ((i>>len)+(j>>len)&1) vis[i][j][op]=1;
		}
	}
	rep (i,0,2*n-1) rep (j,0,2*n-1)
		if (!vis[i][j][0]&&!vis[i][j][1]){
			printf("%d %d\n",i,j);
			if (++cnt==n*n) exit(0);
		}
	return 0;
}