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=1050;
int gi(){
    char ch=getchar(); int x=0,q=0;
    while(ch<'0'||ch>'9') q=ch=='-'?1:q,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return q?-x:x;
}
int f[N][N],n,d1,d2,cnt;
void solve(int d){
    int t=0;
    while(d%4==0)
	d/=4,++t;
    if (d%2)
	for (int i=0; i<2*n; ++i)
	    for (int j=0; j<2*n; ++j)
		f[i][j]|=(i>>t&1)^(j>>t&1);
    else
	for (int i=0; i<2*n; ++i)
	    for (int j=0; j<2*n; ++j)
		f[i][j]|=(i>>t&1);
    return;
}
int main(){
    cin>>n>>d1>>d2,cnt=n*n;
    solve(d1),solve(d2);
    for (int i=0; i<2*n; ++i)
	for (int j=0; j<2*n; ++j)
	    if (cnt&&!f[i][j])
		--cnt,cout<<i<<" "<<j<<endl;
    return 0;
}
