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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdio>
#define N 604
using namespace std;

int n, m, a[N][N], num[6];

inline void work(int d, int op) {
    int c = 0;
    while (d % 4 == 0) d >>= 2, ++c;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j) {
            if ((d & 1) == 0) a[i][j] = op * a[i][j] + ((i >> c) & 1);
            else a[i][j] = op * a[i][j] + (((i >> c) + (j >> c)) & 1);
        }
}

int main() {
#ifdef whyqx
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif

    int d1, d2;
    cin >> n >> d1 >> d2, m = n * n, n <<= 1;
    work(d1, 0), work(d2, 2);

    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
            ++num[a[i][j]];

    int id = 0;
    while (num[id] < m) ++id;

    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
            if (m && a[i][j] == id) printf("%d %d\n", i, j), --m;

    return 0;
}
