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 _=600+7;
int n,D[2],num[2],cnt[2][2];
bool col[_][_][2],t1,t2;
void init(){
  scanf("%d%d%d",&n,&D[0],&D[1]);
  //printf("%d %d %d\n",n,D[0],D[1]);
  for(int i=0;i<=1;i++)
    while(!(D[i]&1)){
      num[i]++;
      D[i]>>=1;
      //printf("num[%d]: %d\n",i,num[i]);
    }
}
void dye(int t){
  for(int i=0;i<2*n;i++)
    for(int j=0;j<2*n;j++)
      if(!(num[t]&1)) col[i][j][t]=((i+j)>>(num[t]>>1))&1;
      else col[i][j][t]=(i>>(num[t]>>1))&1;
}
void calc(){
  for(int i=0;i<2*n;i++)
    for(int j=0;j<2*n;j++)
      cnt[col[i][j][0]][col[i][j][1]]++;
  for(int i=0;i<=1;i++)
    for(int j=0;j<=1;j++)
      if(cnt[i][j]>cnt[t1][t2]){ t1=i; t2=j; }
}
void print(){
  int cnt=0;
  for(int i=0;i<2*n;i++)
    for(int j=0;j<2*n;j++)
      if(col[i][j][0]==t1&&col[i][j][1]==t2){
	printf("%d %d\n",i,j);
	cnt++;
	if(cnt==n*n) return;
      }
}
int main(){
  /*
#ifndef ONLINE_JUDGE
  freopen("x.in","r",stdin);
  freopen("x.out","w",stdout);
#endif
  */
  init();
  //printf("%d %d\n",num[0],num[1]);
  dye(0);
  dye(1);
  calc();
  print();
  return 0;
}
