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;
using i64 = long long;
#define rep(i,s,e) for(int (i) = (s);(i) <= (e);(i)++)

#define int long long

int N,D[2];
int bit[1010][1010];
signed main(){
  cin >> N >> D[0] >> D[1];
  for(int c = 0;c < 2;c++){
    int d = D[c];
    int p = 0;
    while(d % 2 == 0){
      p++;
      d /= 2;
    }
    if(p & 1){
      // ボーダー
      int h = (1 << (p / 2));
      rep(i,0,N * 2 - 1) rep(j,0,N * 2 - 1){
        if((j % (h * 2)) >= h) bit[i][j] = 1;
      }
    }
    else{
      int h = (1 << (p / 2));
      rep(i,0,N * 2 - 1) rep(j,0,N * 2 - 1){
        if(((i % (h * 2)) >= h) xor ((j % (h * 2)) >= h)){
          bit[i][j] = 1;
        }
      }
    }
  }

  int cou = 0;
  rep(i,0,N * 2 - 1) rep(j,0,N * 2 - 1){
    if(!bit[i][j]){
      cou++;
      cout << i << " " << j << endl;
    }
    if(cou == N * N) return 0;
  }
  if(cou != N * N){
    int a = 0;
    cout << 4 / a << endl;
  }
}
