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
/* Your prologue will be the story. */
/* Copyright by HenryLi */
#include<iostream>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<vector>
#include<set>
#include<sstream>
#include<stack>
#include<ctime>
#include<cmath>
#include<cctype>
#include<climits>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<iomanip>
#include<bitset>
#include<complex>
using namespace std;
template <class _T> inline void read(_T &_x) {
	int _t; bool _flag=false;
	while((_t=getchar())!='-'&&(_t<'0'||_t>'9')) ;
	if(_t=='-') _flag=true,_t=getchar(); _x=_t-'0';
	while((_t=getchar())>='0'&&_t<='9') _x=_x*10+_t-'0';
	if(_flag) _x=-_x;
}
typedef long long LL;
int n, d1, d2, c1, c2;
bool check(int a, int b, int c) {
	if (c & 1) return (a >> (c / 2)) % 2 == 0;
	return (a >> (c / 2)) % 2 == (b >> (c / 2)) % 2;
}
int main() {
	//freopen("D.in","r",stdin);
	//freopen("D.out","w",stdout);
	read(n), read(d1), read(d2);
	while (d1 % (1 << (c1 + 1)) == 0) ++c1;
	while (d2 % (1 << (c2 + 1)) == 0) ++c2;
	int tot = 0;
	for (int i = 0; i < 2 * n && tot < n * n; ++i) {
		for (int j = 0; j < 2 * n && tot < n * n; ++j) {
			if (check(i, j, c1) && check(i, j, c2)) {
				cout << i << ' ' << j << endl;
				++tot;
			}
		}
	}
	return 0;
}

