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>
using namespace std;

#define int long long
const double PI = 3.14159265358979323846;
typedef vector<int> vint;
typedef pair<int, int> pint;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

int N, D1, D2;

signed main() {
    cin >> N >> D1 >> D2;
    int p1 = 0, p2 = 0;
    while (D1 % 2 == 0) {D1 /= 2; p1++;}
    while (D2 % 2 == 0) {D2 /= 2; p2++;}
    int n = 0;
    for (int i = 0; i < 2 * N; i++) {
        for (int j = 0; j < 2 * N; j++) {
            bool sw1, sw2;

            if (p1 % 2 == 0)sw1 = (i / (1 << (p1 / 2)) % 2 == j / (1 << (p1 / 2)) % 2);
            else sw1 = (i / (1 << (p1 / 2)) % 2 == 0);

            if (p2 % 2 == 0)sw2 = (i / (1 << (p2 / 2)) % 2 == j / (1 << (p2 / 2)) % 2);
            else sw2 = (i / (1 << (p2 / 2)) % 2 == 0);

            if (sw1 && sw2) {cout << i << " " << j << endl; n++;}
            if (n == N * N)return 0;
        }
    }
}