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;
    template <typename T> void read(T &t) {
    	t=0; char ch=getchar(); int f=1;
    	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
    	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
    }
    const int maxn=610;
    int n,m,x,lg;
    bool c[maxn][maxn];
    void solve() {
    	lg=0;
    	read(x);
    	while (x%4==0)
    		x/=4,lg++;
    	if (x%2==1) {
    		for (int i=0;i<2*n;i++)
    		for (int j=0;j<2*n;j++)
    			c[i][j]|=((i>>lg)+(j>>lg))&1;
    	} else {
    		for (int i=0;i<2*n;i++)
    			if (i&(1<<lg))
    				for (int j=0;j<2*n;j++) c[i][j]=1;
    	}
    }
    int main() {
    	//freopen("1.txt","r",stdin);
    	read(n);
    	solve();
    	solve();
    	int tot=0;
    	for (int i=0;i<2*n;i++)
    	for (int j=0;j<2*n;j++)
    		if (!c[i][j]) {
    			printf("%d %d\n",i,j);
    			tot++;
    			if (tot==n*n) return 0;
    		}
        return 0;
    }