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;

void col(int n,int m,bool p[][605]) {
  int d=0;
  while (m%(1<<(2*(d+1)))==0) d++;
  d=(1<<d);
  int v=(m/(d*d))%4;
  if (v==0||v==3) return;
  if (v==1) {
  	for(int i=0;i<n;i++)
  	  for(int j=0;j<n;j++) p[i][j]=(((i/d)+(j/d))&1);
  }
  else {
  	for(int i=0;i<n;i++)
  	  for(int j=0;j<n;j++) p[i][j]=((i/d)&1);
  }
}

bool f[605][605],g[605][605];

int main() {
  int n,D1,D2;
  scanf("%d%d%d",&n,&D1,&D2);
  col(n<<1,D1,f);
  col(n<<1,D2,g);
  int sz[2][2]={0};
  for(int i=0;i<(n<<1);i++)
    for(int j=0;j<(n<<1);j++) sz[f[i][j]][g[i][j]]++;
  int id1=0,id2=0;
  for(int i=0;i<2;i++)
    for(int j=0;j<2;j++)
      if (sz[i][j]>sz[id1][id2]) {
      	id1=i;
      	id2=j;
	  }
  int cnt=0;
  for(int i=0;i<(n<<1);i++) {
    for(int j=0;j<(n<<1);j++)
      if (f[i][j]==id1&&g[i][j]==id2) {
      	printf("%d %d\n",i,j);
      	cnt++;
      	if (cnt>=n*n) break;
	  }
	if (cnt>=n*n) break;
  }
  return 0;
}
