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;

typedef long long lint;
typedef long double louble;

const char lf = '\n';

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

namespace ae86
{
	const int bufl = 1<<15;
	
	char buf[bufl],*s=buf,*t=buf;
	
	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}
	
	inline int ty()
	{
		int a=0,b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}
}
using ae86::ty;

const int _ = 303 , __ = _<<1;

int n,nn,ed[__][__]={0};

inline void make(int ps)
{
	int lg=0;
	while(ps%4==0)ps/=4,lg++;
	if(ps%4==1)
	{
		for(int i=1;i<=nn;i++)
			for(int j=1;j<=nn;j++)
				ed[i][j]|=(((i-1)>>lg)+((j-1)>>lg))&1;
	}
	else
	{
		for(int i=1;i<=nn;i++)
			for(int j=1;j<=nn;j++)
				ed[i][j]|=((i-1)>>lg)&1;
	}
}

int main()
{
	ios::sync_with_stdio(0),cout.tie(nullptr);
	
	n=ty(),nn=n+n,make(ty()),make(ty());
	int cnt=n*n;

	for(int i=1;i<=nn && cnt>0;i++)
		for(int j=1;j<=nn && cnt>0;j++)
			if(!ed[i][j])cout<<i-1<<' '<<j-1<<lf,cnt--;
	
	return 0;
}
