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 <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N=610;
int n,m;
bool w[N][N];
inline void solve() {
    int x,d=0;
    cin>>x;
    while(x%4==0) x/=4,++d;
    if(x&1) {
	for(int i=0;i<m;i++)
	    for(int j=0;j<m;j++)
		if(((i>>d)+(j>>d))&1) w[i][j]=1;
    }
    else {
	for(int i=0;i<m;i++)
	    if(i>>d&1)
		for(int j=0;j<m;j++) w[i][j]=1;
    }
}
int main() {
    cin>>n,m=n+n;
    solve(),solve(),n*=n;
    for(int i=0;i<m;i++)
	for(int j=0;j<m;j++)
	    if(!w[i][j]) {
		printf("%d %d\n",i,j);
		if(!--n) return 0;
	    }
    return 0;
}