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
//made by Sert
#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9 + 34;
const int N = (int)1e6 + 34;

typedef long long ll;

set<pair<int, int>> s;

void del(int x, int y) {
    pair<int, int> p = make_pair(x, y);
    if (s.find(p) != s.end()) s.erase(p);
}

void go(int n, int mult, int d) {
    if (d % 4 == 3) return;
    if (d % 4 == 0) {
        go(n, mult * 2, d / 4);
    }
    for (int i = 0; i < n * 2; i++) {
        for (int j = 0; j < n * 2; j++) {
            int y = i / mult;
            int x = j / mult;
            if (d % 4 == 1 && (x + y) % 2 == 1) del(j, i);
            if (d % 4 == 2 && x % 2 == 1) del(j, i);
        }
    }
}

void solve() {
    int n, d1, d2;
    cin >> n >> d1 >> d2;
    s.clear();
    for (int i = 0; i < n * 2; i++) {
        for (int j = 0; j < n * 2; j++) {
            s.insert(make_pair(i, j));
        }
    }
    go(n, 1, d1);
    go(n, 1, d2);
    auto it = s.begin();
    for (int i = 0; i < n * n; i++) {
        cout << it->first << " " << it->second << "\n";
        it++;
    }
}

int main() {
#ifdef SERT
    freopen("a.in", "r", stdin);
    //freopen("a.out", "w", stdout);
    //solve();
    //solve();
    //solve();
    //solve();
#endif
    solve();
    return 0;
}