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;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;

int s[2],ii[2],jj[2];
bool f[2];

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  
  int N,D[2]; cin>>N>>D[0]>>D[1];
  rep(i,2){
    while(D[i]%4==0){
      D[i]/=4;
      s[i]++;
    }
  }
  vector<pair<int,int>> V;
  rep(i,2*N){
    rep(k,2) ii[k]=(i>>s[k]);
    rep(j,2*N){
      rep(k,2) jj[k]=(j>>s[k]);
      rep(k,2){
        f[k]=(D[k]%2==1) ? ((ii[k]+jj[k])%2==0) : (ii[k]%2==0) ;
      }
      if(f[0]&&f[1]) V.push_back({i,j});
    }
  }
  rep(i,N*N) cout<<V[i].first<<" "<<V[i].second<<endl;
  return 0;
}
