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<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

vector<P> ans[4], ban[2];
Int d[2];
Int n ;
bool come[2][648][648];
Int color[648][648];
void dfs(int x, int y, int k, int c = 0){
  if(x < 0 || x >= 2*n || y < 0 || y >= 2*n)return;
  if(come[k][x][y])return;
  come[k][x][y] = true;
  color[x][y] +=  c << k;
  for(auto b:ban[k]){
    int nx = x + b.first, ny = y + b.second;
    dfs(nx, ny, k, !c);
  }
}

int main(){
  cin >> n >> d[0] >> d[1];
  for(int i = 0;i < 2*n;i++){
    for(int j = 0;j < 2*n;j++){
      for(int k = 0;k < 2;k++){
	if(i * i + j * j == d[k]){
	  ban[k].push_back(P(i,j));
	  ban[k].push_back(P(i,-j));
	  ban[k].push_back(P(-i,j));
	  ban[k].push_back(P(-i,-j));
	}
      }
    }
  }
  
  for(int i = 0;i < 2*n;i++){
    for(int j = 0;j < 2*n;j++){
      dfs(i, j, 0);
      dfs(i, j, 1);
    }
  }

  for(int i = 0;i < 2*n;i++){
    for(int j = 0;j < 2*n;j++){
      ans[color[i][j]].push_back(P(i,j));
    }
  }
  for(int i = 0;i < 4;i++){
    if(ans[i].size() < n*n)continue;
    for(int j = 0;j < n*n;j++){
      P point = ans[i][j];
      cout << point.first << " " << point.second << endl;
    }
    return 0;
  }
  return 0;
}