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;
#define N 610
#define pb push_back
#define pii pair<int,int>
#define mkp(a,b) make_pair(a,b)
vector <pii> v[2];
int col[2][N][N];
inline void init(int w,int d) {
    for (int i=0;i<=d;i++) {
        int j=sqrt(d-i*i);
        if (i*i+j*j!=d) continue;
        v[w].pb(mkp(i,j)),v[w].pb(mkp(-i,-j));
        v[w].pb(mkp(-i,j)),v[w].pb(mkp(i,-j));
    }
}
inline void dfs(int x,int y,int c,int w,int n) {
    col[w][x][y]=c;
    for (auto &d:v[w]) {
        int _x=x+d.first,_y=y+d.second;
        if (_x<0 || 2*n<=_x || _y<0 || 2*n<=_y) continue;
        if (!col[w][_x][_y]) dfs(_x,_y,3-c,w,n);
    }
}
int main() {
    int n,d1,d2,cnt=0; 
    cin>>n>>d1>>d2;
    init(0,d1),init(1,d2);
    for (int i=0;i<=2*n-1;i++)
        for (int j=0;j<=2*n-1;j++) {
            if (!col[0][i][j]) dfs(i,j,1,0,n);
            if (!col[1][i][j]) dfs(i,j,1,1,n);
        }
    for (int i=0;i<=2*n-1;i++)
        for (int j=0;j<=2*n-1;j++)
            if (col[0][i][j]==1 && col[1][i][j]==1) {
                printf("%d %d\n",i,j);
                if ((++cnt)==n*n) return 0;
            }
    return 0;
}