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 <fstream>
#include <vector>

using namespace std;
const int nmax=305;
vector< pair<int,int> > ap[5][5],offsets;
int col[2][2*nmax][2*nmax];
int n,d1,d2,i,j,k;
void dfs(int wh,int li,int ci)
{
    int lf,cf;
    for(int i=0;i<offsets.size();i++)
    {
        lf=li+offsets[i].first;
        cf=ci+offsets[i].second;
        if(lf>=0&&lf<2*n&&cf>=0&&cf<2*n&&(!col[wh][lf][cf]))
       {
          col[wh][lf][cf]=(3-col[wh][li][ci]);
          dfs(wh,lf,cf);
       }
    }
}
void solve(int wh,int d)
{
    offsets.clear();
    for(i=-2*n;i<=2*n;i++)
        for(j=-2*n;j<=2*n;j++)
           if(i*i+j*j==d)
               offsets.push_back({i,j});
    for(int i=0;i<2*n;i++)
        for(int j=0;j<2*n;j++)
         if(!col[wh][i][j])
            {
                col[wh][i][j]=1;
                dfs(wh,i,j);
            }
}
int main()
{
   // freopen("data.in","r",stdin);
    ios_base::sync_with_stdio(false);
    cin>>n>>d1>>d2;
    solve(0,d1);
    solve(1,d2);
    for(i=0;i<2*n;i++)
        for(j=0;j<2*n;j++)
         ap[col[0][i][j]][col[1][i][j]].push_back({i,j});
    for(i=1;i<=2;i++)
        for(j=1;j<=2;j++)
          if(ap[i][j].size()>=n*n)
    {
        for(k=0;k<n*n;k++)
            cout<<ap[i][j][k].first<<' '<<ap[i][j][k].second<<'\n';
        return 0;
    }
    return 0;
}
