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 <cstdio>
#include <vector>

int color[600][600];
int N;

void calccolor(int D,int s){
  int scale=1;
  while(D%4==0){
    scale*=2;
    D/=4;
  }
  if(D%2==0){
    for(int i=0;i<N*2;i++){
      for(int j=0;j<N*2;j++){
	color[i][j]|=(((i/scale)&1)<<s);
      }
    }
  }else{
    for(int i=0;i<N*2;i++){
      for(int j=0;j<N*2;j++){
	color[i][j]|=((((i/scale)^(j/scale))&1)<<s);
      }
    }
  }
}

std::vector<std::pair<int,int> > points[4];

int main(){
  int D1,D2;
  scanf("%d %d %d",&N,&D1,&D2);
  calccolor(D1,0);
  calccolor(D2,1);
  for(int i=0;i<N*2;i++){
    for(int j=0;j<N*2;j++){
      int d=i*i+j*j;
      points[color[i][j]].emplace_back(i,j);
      //printf("%d",color[i][j]);
    }
    //printf("\n");
  }
  for(int i=0;i<4;i++){
    if(points[i].size()>=N*N){
      std::swap(points[i],points[0]);
      break;
    }
  }
  for(int i=0;i<N*N;i++){
    printf("%d %d\n",points[0][i].first,points[0][i].second);
  }
  return 0;
}
