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;

int n, d1, d2;
char col[601][601];

int color(int x, int y, int d)
{
  if (d % 2 == 1) {
    return (x + y) % 2;
  }
  if (d % 4 == 2) {
    return x % 2;
  }
  return color(x/2, y/2, d/4);
}

int main()
{
  cin >> n >> d1 >> d2;

  for (int x = 0; x < 2*n; x++)
    for (int y = 0; y < 2*n; y++)
      col[x][y] = 0;

  for (int x = 0; x < 2*n; x++) {
    for (int y = 0; y < 2*n; y++) {
      col[x][y] = color(x, y, d1) | (color(x, y, d2) << 1);
    }
  }

  int c[4] = {};
  for (int x = 0; x < 2*n; x++)
    for (int y = 0; y < 2*n; y++)
      c[(int)col[x][y]]++;

  int v = max_element(c, c+4) - c;
  int k = 0;
  for (int x = 0; x < 2*n; x++)
    for (int y = 0; y < 2*n; y++) {
      if (k == n*n) return 0;
      if (v == col[x][y]) {
        cout << x << " " << y << endl;
        k++;
      }
    }

  return 0;
}
