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<cmath> 

const int Maxv = 605; 
int Map[Maxv][Maxv][2], d[5], n, m, cnt; 
bool vis[Maxv][Maxv][2];

inline void dfs(int x, int y, int a, int b){
    if (x < 0 || x > m || y < 0 || y > m || vis[x][y][a]) {
        return; 
    }
    vis[x][y][a] = true; 
    Map[x][y][a] = b; 
    for (int i = 0; i * i <= d[a]; i++) {
        int j = sqrt(d[a] - i * i); 
        if (i * i + j * j != d[a]) {
            continue; 
        }
        dfs(x + i, y + j, a, 3 - b); 
        dfs(x - i, y + j, a, 3 - b); 
        dfs(x + i, y - j, a, 3 - b); 
        dfs(x - i, y - j, a, 3 - b); 
    }
}

int main(){
    scanf("%d%d%d", &n, &d[0], &d[1]); 
    m = 2 * n - 1; 
    for (int i = 0; i <= m; i++) {
        for (int j = 0; j <= m; j++) {
            if (!vis[i][j][0]) {
                dfs(i, j, 0, 1); 
            }
            if (!vis[i][j][1]) {
                dfs(i, j, 1, 1); 
            }
        }
    }
    for (int i = 0; i <= m; i++) {
        for (int j = 0; j <= m; j++) {
            if (Map[i][j][0] == 1 && Map[i][j][1] == 1) {
                printf("%d %d\n", i, j); 
                cnt++; 
                if (cnt == n * n) {
                    return 0; 
                }
            }
        }
    }
}