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 rep(i,a,b) for(int i = (a); i < (b); ++i)
#define rrep(i,a,b) for(int i = (b); i --> (a);)
#define all(v) (v).begin(),(v).end()
#define trav(x,v) for(auto &x : v)
#define sz(v) int(v.size())
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);

	int n, d1, d2;
	cin >> n >> d1 >> d2;

	vector<vi> c(2*n, vi(2*n));

	function<bool(int,int,int)> isgood = [&](int k, int i, int j){
		if(k == 0) return (i+j)%2 == 0;
		if((i+j)%2) ++j;
		int p = (i+j)/2;
		int q = (i-j)/2;
		return isgood(k-1, p, q);
	};

	auto f = [&](int d, int dif){
		int v2 = 0;
		while(d%2 == 0){
			d /= 2;
			 v2++;
		}
		rep(i,0,2*n) rep(j,0,2*n) c[i][j] += dif * isgood(v2, i, j);
	};
	f(d1, 1);
	f(d2, 2);

	vi cnt(4);

	rep(i,0,2*n) rep(j,0,2*n) cnt[c[i][j]]++;

	rep(t,0,4){
		if(cnt[t] >= n*n){
			int cur = 0;
			rep(i,0,2*n) rep(j,0,2*n) if(c[i][j] == t){
				cout << i << ' ' << j << endl;
				if(++cur == n*n) return 0;
			}
		}
	}
}