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>
#define meow(args...) fprintf(stderr, args)
typedef unsigned u32;
typedef long long s64;
typedef unsigned long long u64;
template<class T1, class T2> inline bool cmin(T1 &a, const T2 &b) {return b<a?(a=b, true):false;}
template<class T1, class T2> inline bool cmax(T1 &a, const T2 &b) {return a<b?(a=b, true):false;}

int n, c[600][600];
void color(int alp, int id) {
	for(int i=0; i<n; ++i)
		for(int j=0; j<n; ++j)
			c[i][j]+=((i+(~alp&1)*j)>>(alp/2)&1)<<id;
}
int main() {
	int d1, d2, cnt[4]={};
	assert(scanf("%d%d%d", &n, &d1, &d2)==3);
	n*=2;
	color(__builtin_ctz(d1), 0);
	color(__builtin_ctz(d2), 1);
	for(int i=0; i<n; ++i)
		for(int j=0; j<n; ++j) ++cnt[c[i][j]];
	int maj=std::max_element(cnt, cnt+4)-cnt, cur=0;
	for(int i=0; i<n; ++i)
		for(int j=0; j<n; ++j)
			if(c[i][j]==maj) {
				printf("%d %d\n", i, j);
				if(++cur==n*n/4) return 0;
			}
	return 0;
}
