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>
#define RI register int
typedef long long LL;

#define FILEIO(name) freopen(name".in", "r", stdin), freopen(name".out", "w", stdout);

using namespace std;

char buf[1000000], *p1 = buf, *p2 = buf;
inline char gc() {
  if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);
  return p1 == p2 ? EOF : *(p1++);
}
template <class T> inline void read(T &n) {
  n = 0; RI ch = gc(), f;
  while ((ch < '0' || ch > '9') && (ch != '-')) ch = gc();
  f = (ch == '-' ? ch = gc(), -1 : 1);
  while (ch >= '0' && ch <= '9') n = n * 10 + (ch ^ 48), ch = gc();
  n *= f;
}

int col[605][605];

void Solve(int n, int k) {
  int t = 0;
  while (!(k % 4)) k >>= 2, ++t;
  for (RI i = 0; i < 2 * n; ++i)
    for (RI j = 0; j < 2 * n; ++j) {
      if (k % 4 == 1) {
        if (((i >> t) + (j >> t)) & 1)
          col[i][j] = 1;
      }
      else
        if ((i >> t) & 1)
          col[i][j] = 1;
    }
  // if t % 2 == 0 then t^2 % 4 = 0
  // if t % 2 == 1 then t^2 % 4 = 1
  // so if k = a^2 + b^2 then k % 4 != 3
}

int main() {

//  FILEIO("a");

  int n, d1, d2; read(n), read(d1), read(d2);
  Solve(n, d1), Solve(n, d2);
  int cnt = 0;
  for (RI i = 0; i < 2 * n && cnt < n * n; ++i)
    for (RI j = 0; j < 2 * n && cnt < n * n; ++j)
      if (!col[i][j])
        printf("%d %d\n", i, j), ++cnt;

  cerr << (double)(clock()) / CLOCKS_PER_SEC * 1000 << "ms" << endl;

  return 0;
}

// created by Daniel yuan