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 mset(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int MAXN = 1005;
 
template <typename T> inline void read(T &a)
{
    T x=0, flag=1;
    char ch = getchar();
    while(!isdigit(ch)){
        if(ch == '-') flag = -1;
        ch = getchar();
    }
    while(isdigit(ch)){
        x = (x<<1) + (x<<3) + ch - '0';
        ch = getchar();
    }
    a = flag * x;
}

int n, d1, d2;
bool vis[MAXN][MAXN];

void build(int d)
{
	int t = 0;
	while(d % 4 == 0){
		d /= 4;
		++t;
	}
	if(d%4 == 1){
		for(int i=0;i<n;++i){
			for(int j=0;j<n;++j){
				if(((i>>t) + (j>>t)) & 1) vis[i][j] = 1;
			}
		}
	}else{
		for(int i=0;i<n;++i){
			for(int j=0;j<n;++j){
				if((i>>t) & 1) vis[i][j] = 1;
			}
		}
	}
}

void solve()
{
	read(n); read(d1); read(d2);
	int cnt = n * n, out = 0;
	n <<= 1;
	build(d1); build(d2);
	for(int i=0;i<n;++i){
		for(int j=0;j<n;++j){
			if(!vis[i][j]){
				printf("%d %d\n", i, j);
				if(++out == cnt) return;
			}
		}
	}
}

int main()
{
	solve();
	return 0;
}
