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>

using namespace std;
const int maxn=6e2+45;
#define INF 0x3f3f3f3f;
#define ll long long int
#define charmax(x,y) x=max(x,y)
#define charmin(x,y) x=min(x,y)
#define clr(x,p) memset(x,p,sizeof x)
int n,d1,d2;int w;
int a[maxn][maxn];
int vis[maxn][maxn];
void work(int d){
    int pos=0;
    while(d%4==0){
        d/=4;
        pos++;
    }
    if(d%2==1){
        for(int i=0;i<w;i++){
            for(int j=0;j<w;j++){
                if(((i>>pos)+(j>>pos))&1){
                    a[i][j]=1;
                }
            }
        }
    }
    else{
        for(int i=0;i<w;i++){
            for(int j=0;j<w;j++){
                if((i>>pos)&1) a[i][j]=1;
            }
        }
    }
}
int main()
{
    clr(a,0);clr(vis,0);
    scanf("%d %d %d",&n,&d1,&d2);
    w=2*n;
    work(d1);work(d2);
    int cs=0;
    for(int i=0;i<w;i++){
        for(int j=0;j<w;j++){
            if(cs==n*n) break;
            if(a[i][j]==0){
                printf("%d %d\n",i,j);
                cs++;
            }
        }
    }
    return 0;
}
