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>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define FORR(i, n) for(ll i = (ll)n - 1LL; i >= 0LL; i--)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) begin(x), end(x)
using namespace std;
using ll = long long;
template <typename T> using V = vector<T>;
constexpr ll Mod = 998244353;
constexpr ll mod = 1e9 + 7;
constexpr ll inf = 1LL << 60;
const double PI = acos(-1);
template <typename T> constexpr bool chmax(T &a, const T &b) {
    if(a >= b) return false;
    a = b;
    return true;
}
template <typename T> constexpr bool chmin(T &a, const T &b) {
    if(a <= b) return false;
    a = b;
    return true;
}

/*-------------------------------------------*/

bool check(int i, int j, int d) {
    int u = __builtin_ctz(d);
    return !((u & 1 ? i : i ^ j) >> (u / 2) & 1);
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int n, d1, d2, cnt = 0;
    cin >> n >> d1 >> d2;
    rep(i, 2 * n) rep(j, 2 * n) {
        if(check(i, j, d1) && check(i, j, d2)) {
            cout << i << " " << j << "\n";
            if(++cnt >= n * n) return 0;
        }
    }

    return 0;
}