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>
// #pragma GCC optimize ("O3")
// #pragma GCC target ("sse4")
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;

#define REP(i,n) for(int i=0;i<(n);++i)
#define FOR(i,a,b) for (int i=(a); i<(b); ++i)
#define FORD(i,a,b) for (int i=(a)-1; i>=(b); --i)

#define pb push_back
#define mp make_pair
#define st first
#define nd second

const int MOD = 1000000007;

bool T[605][605];

int gcd(int a, int b) {
  return b ? gcd(b, a%b) : a;
}

int main() {
  // ios_base::sync_with_stdio(0);

  int N, D1, D2;
  scanf("%d%d%d", &N, &D1, &D2);
  vector<int> Ds = {D1, D2};

  REP(i,2*N)REP(j,2*N) T[i][j] = true;
  for (auto d: Ds) {
    int p = 1;
    while (d % (4 * p * p) == 0) p *= 2;
    if ((d/p/p) % 4 == 1) {
      REP(i,2*N)REP(j,2*N) if ((i/p) % 2 != (j/p) % 2) T[i][j] = false;
    } else {
      REP(i,2*N)REP(j,2*N) if ((i/p) % 2 == 1) T[i][j] = false;
    }
  }
  //
  // REP(i,2*N) {
  //   REP(j,2*N) printf(T[i][j] ? "#" : ".");
  //   printf("\n");
  // }

  vector<PII> good;
  REP(i,2*N)REP(j,2*N) if (T[i][j]) {
    good.pb({i,j});
  }

  assert(good.size() >= N*N);
  REP(i,N*N) printf("%d %d\n", good[i].st, good[i].nd);
}
