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;

char BUF[1 << 21] , *buf = BUF;
void putch(char c){*(buf++) = c;}
void putint(int x){if(!(x / 10)){putch('0' + x); return;} putint(x / 10); putch('0' + x % 10);}
void flush(){fwrite(BUF , 1 , buf - BUF , stdout);}

bool vis[603][603]; int N , D1 , D2;
void color(int dist){
	int now = 0; while(!(dist & 3)){dist >>= 2; ++now;}
	if(dist % 4 == 3) return;
	for(int i = 0 ; i < 2 * N ; ++i)
		for(int j = 0 ; j < 2 * N ; ++j)
			vis[i][j] |= ((dist & 2) && (i >> now & 1)) || ((dist & 1) && (((i >> now) + (j >> now)) & 1));
}

int main(){
	cin >> N >> D1 >> D2; color(D1); color(D2); int num = 0;
	for(int i = 0 ; i < 2 * N ; ++i)
		for(int j = 0 ; j < 2 * N ; ++j)
			if(!vis[i][j] && ++num <= N * N){putint(i); putch(' '); putint(j); putch('\n');}
	flush(); return 0;
}