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;
int n,col1[605][605],col2[605][605];
vector<pair<int,int>>s[2];
void Init(int d,vector<pair<int,int>>&v){
    for(int i=0;i<=d;i++){
        int j=sqrt(d-i*i);
        if(i*i+j*j!=d)continue;
        v.emplace_back(+i,+j);
        v.emplace_back(+i,-j);
        v.emplace_back(-i,+j);
        v.emplace_back(-i,-j);
    }
}
void dfs(int x,int y,int c,vector<pair<int,int>>&v,int col[][605]){
    col[x][y]=c;
    for(auto d:v){
        int dx,dy;tie(dx,dy)=d;
        dx+=x,dy+=y;
        if(dx<0||dy<0||dx>=2*n||dy>=2*n)continue;
        if(!col[dx][dy])
            dfs(dx,dy,-c,v,col);
    }
}
void Solve(vector<pair<int,int>>&v,int col[][605]){
    for(int i=0;i<2*n;i++)
        for(int j=0;j<2*n;j++)
            if(!col[i][j])
                dfs(i,j,1,v,col);
}
int main(){
    int d1,d2;
    cin>>n>>d1>>d2;
    Init(d1,s[0]);
    Init(d2,s[1]);
    Solve(s[0],col1);
    Solve(s[1],col2);
    vector<pair<int,int>>ans;
    for(int i=0;i<2*n;i++)
        for(int j=0;j<2*n;j++)
            if(col1[i][j]==col1[0][0]&&col2[i][j]==col2[0][0])
                ans.emplace_back(i,j);
    for(int i=0;i<n*n;i++)
        cout<<ans[i].first<<" "<<ans[i].second<<"\n";
    return 0;
}