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
//感谢Factorio
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <set>
#include <queue>
#include <map>
using namespace std;
int n,D1,D2;
int a[1005][1005];
void Doit(int dist){
    int c=0;
    while (dist%4==0){
        dist/=4;
        c++;
    }
    if (dist%2==0){
        for (int i=0;i<2*n;i++){
            for (int j=0;j<2*n;j++){
                if ((i>>c)&1){
                    a[i][j]=1;
                }
            }
        }
    }
    else{
        for (int i=0;i<2*n;i++){
            for (int j=0;j<2*n;j++){
                if (((i>>c)+(j>>c))&1){
                    a[i][j]=1;
                }
            }
        }
    }
}
int main() {
    cin>>n>>D1>>D2;
    Doit(D1);
    Doit(D2);
    int cnt=0;
    for (int i=0;i<2*n;i++){
        for (int j=0;j<2*n;j++){
            if (a[i][j]==0){
                cnt++;
                cout<<i<<" "<<j<<endl;
                if (cnt==n*n){
                    return 0;
                }
            }
        }
    }
    return 0;
}