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 <iostream>
using namespace std;

int N, cnt[610][610];

void fill(int d){
	int a=1, b=d;
	while(b%4==0){
		a *= 2;
		b /= 4;
	}
	for(int i=0; i<N*2; ++i){
		for(int j=0; j<N*2; ++j){
			if(b%4==3) ++cnt[i][j];
			else if(b%4==1 && (i/a+j/a)%2==0) ++cnt[i][j];
			else if(b%4==2 && (i/a)%2==0) ++cnt[i][j];
		}
	}
}

int main() {
	int D1, D2;
	cin >> N >> D1 >> D2;
	fill(D1);
	fill(D2);
	int sum=0;
	for(int i=0; i<N*2; ++i){
		for(int j=0; j<N*2; ++j){
			if(cnt[i][j]==2){
				cout << i << ' ' << j << endl;
				++sum;
				if(sum==N*N) return 0;
			}
		}
	}
	return 0;
}