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 <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <map>
#include <set>
using namespace std;
const int N=660,M=100010;
const int MAX=2e9;
const long long Mod=1e9+7;
int vis[N][N];
int n,d1,d2,bd,tot,tar;
void deal(int dis){
    int g=0;
    while(!(dis%4)){
        dis/=4;
        g++;
    }
    if(dis&1){
        for(int i=0;i<bd;i++){
            for(int j=0;j<bd;j++){
                if((i>>g)+(j>>g)&1) vis[i][j]=1;
            }
        }
    }
    else{
        for(int i=0;i<bd;i++){
            for(int j=0;j<bd;j++){
                if((i>>g)&1) vis[i][j]=1;
            }
        }
    }
    return ;
}
int main(){
    //freopen("problem.in","r",stdin);
    //freopen("problem.out","w",stdout);
    scanf("%d%d%d",&n,&d1,&d2);
    bd=n<<1;
    deal(d1);
    deal(d2);
    tar=n*n;
    for(int i=0;i<bd;i++){
        for(int j=0;j<bd;j++){
            if(!vis[i][j]){
                printf("%d %d\n",i,j);
                tot++;
            }
            if(tot==tar) return 0;
        }
    }
    return 0;
}