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 <algorithm>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#define R register
#define IN inline
#define W while
#define gc getchar()
#define MX 610
bool vis[MX][MX];
int siz, d1, d2, bd;
void deal(R int dis)
{
    int tim = 0;
    W (!(dis % 4)) dis /= 4, ++tim;
    if(dis & 1)
    {
        for (R int i = 0; i < bd; ++i)
        for (R int j = 0; j < bd; ++j)
        if((i >> tim) + (j >> tim) & 1) vis[i][j] = true;
    }
    else
    {
        for (R int i = 0; i < bd; ++i)
        for (R int j = 0; j < bd; ++j)
        if((i >> tim) & 1) vis[i][j] = true;
    }
}
int main(void)
{
    scanf("%d%d%d", &siz, &d1, &d2);
    bd = siz << 1; deal(d1), deal(d2);
    int tar = siz * siz, cnt = 0;
    for (R int i = 0; i < bd; ++i)
    for (R int j = 0; j < bd; ++j)
    {
        if(!vis[i][j]) printf("%d %d\n", i, j), cnt++;
        if(cnt == tar) return 0;
    }
}