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<cstdio>
#define F(i,a) for(int i=0;i<a;++i)
int n,N,D1,D2;
int i1[600],j1[600],t1;
int i2[600],j2[600],t2;
bool v1[600][600], v2[600][600];
bool c1[600][600], c2[600][600];
int s[4];
void DFS1(int x,int y,bool c){
	if(x<0||y<0||x>=N||y>=N) return;
	if(v1[x][y]) return;
	v1[x][y]=1, c1[x][y]=c;
	F(k,t1){
		DFS1(x+i1[k],y+j1[k],!c);
		DFS1(x+i1[k],y-j1[k],!c);
		DFS1(x-i1[k],y+j1[k],!c);
		DFS1(x-i1[k],y-j1[k],!c);
	}
}
void DFS2(int x,int y,bool c){
	if(x<0||y<0||x>=N||y>=N) return;
	if(v2[x][y]) return;
	v2[x][y]=1, c2[x][y]=c;
	F(k,t2){
		DFS2(x+i2[k],y+j2[k],!c);
		DFS2(x+i2[k],y-j2[k],!c);
		DFS2(x-i2[k],y+j2[k],!c);
		DFS2(x-i2[k],y-j2[k],!c);
	}
}
int main(){
	scanf("%d%d%d",&n,&D1,&D2); N=n+n;
	F(i,N) if(i*i<=D1)
		F(j,N) if(i*i+j*j==D1)
			i1[t1]=i, j1[t1]=j, t1++;
	F(i,N) if(i*i<=D2)
		F(j,N) if(i*i+j*j==D2)
			i2[t2]=i, j2[t2]=j, t2++;
	F(i,N) F(j,N) DFS1(i,j,0);
	F(i,N) F(j,N) DFS2(i,j,0);
	F(i,N) F(j,N)
		++s[c1[i][j]*2+c2[i][j]];
	F(k,4) if(s[k]>=n*n){
		int cnt=0;
		F(i,N) F(j,N) if(c1[i][j]*2+c2[i][j]==k&&cnt<n*n)
			++cnt, printf("%d %d\n",i,j);
		return 0;
	}
}