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;
const int N=666;
int vis[2][N][N];
int n,d1,d2,m;
vector<pair<int,int> >v[2];
void dfs(int opt,int x,int y,int st)
{
    if(x<0||x>=m||y<0||y>=m||vis[opt][x][y]) return;
    vis[opt][x][y]=st;
    for(int i=0;i<v[opt].size();i++)
        dfs(opt,x+v[opt][i].first,y+v[opt][i].second,st==1?2:1);
}
int main(void)
{
    scanf("%d%d%d",&n,&d1,&d2);
    m=n*2;
    for(int i=0;i*i<=d1;i++)
    {
        int j=sqrt(d1-i*i);
        if(j*j==d1-i*i)
        {
            v[0].push_back({i,j});
            v[0].push_back({-i,j});
            v[0].push_back({i,-j});
            v[0].push_back({-i,-j});
        }
    }
    for(int i=0;i*i<=d2;i++)
    {
        int j=sqrt(d2-i*i);
        if(j*j==d2-i*i)
        {
            v[1].push_back({i,j});
            v[1].push_back({-i,j});
            v[1].push_back({i,-j});
            v[1].push_back({-i,-j});
        }
    }
    for(int i=0;i<m;i++)
        for(int j=0;j<m;j++)
        for(int k=0;k<2;k++)
        if(!vis[k][i][j])
        {
            dfs(k,i,j,1);
        }
    int k=n*n;
    for(int i=0;i<m;i++)
        for(int j=0;j<m;j++)
        if(vis[0][i][j]==1&&vis[1][i][j]==1)
        {
            printf("%d %d\n",i,j);
            k--;
            if(k==0) return 0;
        }
}
