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 <iostream>
#include <vector>

using namespace std;

const int MAX_N = 605;

int evenp (int d) {
  int ans = 0;
  while (d % 2 == 0) {
    ans++;
    d /= 2;
  }
  return ans;
}

bool fits (int x, int y, int d) {
  if (d % 2 == 0) {
    int pd = 1 << (d / 2);
    return (x / pd + y / pd) % 2 == 0;
  } else {
    int pd = 1 << ((d - 1) / 2);
    return (x / pd) % 2 == 0;
  }
}

int main () {
  ios::sync_with_stdio(false);

  int n, d1, d2;
  cin >> n >> d1 >> d2;

  d1 = evenp(d1);
  d2 = evenp(d2);

  vector<pair<int, int>> ans;
  for (int i = 0; i < 2 * n; i++) {
    for (int j = 0; j < 2 * n; j++) {
      if (fits(i, j, d1) && fits(i, j, d2)) {
        ans.push_back(make_pair(i, j));
      }
    }
  }

  while ((int) ans.size() > n * n) {
    ans.pop_back();
  }

  for (auto pr : ans) {
    cout << pr.first << " " << pr.second << '\n';
  }
}
