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;
using Int = long long;
//INSERT ABOVE HERE
Int dp[666][666];
signed main(){
  Int n,d1,d2;
  cin>>n>>d1>>d2;

  memset(dp,0,sizeof(dp));

  auto in=
    [&](Int y,Int x){
      return 0<=y&&y<n*2&&0<=x&&x<n*2;
    };
  auto calc=
    [&](Int d,Int x){
      Int k=1;
      while(d%4==0) d/=4,k*=2;
      for(Int a=0;a<n*2;a+=k)
	for(Int b=((d&1)&((a/k)&1))*k;b<n*2;b+=k*2)
	  for(Int i=0;i<k;i++)
	    for(Int j=0;j<k;j++)
	      if(in(a+i,b+j)) dp[a+i][b+j]|=x;
    };
  
  calc(d1,1);
  calc(d2,2);
  if(0){
    for(Int i=0;i<n*2;i++){
      for(Int j=0;j<n*2;j++)
	cout<<dp[i][j]<<" ";
      cout<<endl;
    }
  }

  Int cnt[4]={};
  for(Int i=0;i<n*2;i++)
    for(Int j=0;j<n*2;j++)
      cnt[dp[i][j]]++;

  for(Int k=0;k<4;k++){
    if(cnt[k]>=n*n){
      Int res=0;
      for(Int i=0;i<n*2;i++){
	for(Int j=0;j<n*2;j++){
	  if(dp[i][j]!=k) continue;
	  cout<<i<<" "<<j<<endl;
	  if(++res==n*n) return 0;
	}
      }
    }
  }
  
  return 0;
}
