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>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <set>
using namespace std;
typedef long long i64;
typedef pair<int, int> pii;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 6e2+5;
int n, d1, d2, c[N][N], tot;
void solve(int d)
{
    int t = 0; while(!(d&3)) ++t, d >>= 2;
    if((d&3) == 1) 
    {
        for(int i = 0; i < n<<1; ++i)
            for(int j = 0; j < n<<1; ++j)
                c[i][j] |= ((i>>t)+(j>>t))&1; 
    }
    else
    {
        for(int i = 0; i < n<<1; ++i)
            for(int j = 0; j < n<<1; ++j)
                c[i][j] |= (i>>t)&1;      
    }
}
int main()
{
    n = read(), d1 = read(), d2 = read(), solve(d1), solve(d2);
    for(int i = 0; i < n<<1; ++i)
        for(int j = 0; j < n<<1; ++j)
            if(!c[i][j]&&++tot <= n*n) printf("%d %d\n", i, j);
    return 0;
}