Problem:
Score: 500 points

Problem Statement

We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.

We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.

The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.

In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

-10^{18} \leq A_i, B_i \leq 10^{18}

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_N B_N

Output

Print the count modulo 1000000007.

Sample Input 1

3
1 2
-1 1
2 -1

Sample Output 1

5

There are five ways to choose the set of sardines, as follows:

The 1-st

The 1-st and 2-nd

The 2-nd

The 2-nd and 3-rd

The 3-rd

Sample Input 2

10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4

Sample Output 2

479

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pi = pair<int, int>;
const int mod = 1000000007;
struct mint
{
    ll x; // typedef long long ll;
    mint(ll x = 0) : x((x % mod + mod) % mod) {}
    mint operator-() const { return mint(-x); }
    mint &operator+=(const mint a)
    {
        if ((x += a.x) >= mod)
            x -= mod;
        return *this;
    }
    mint &operator-=(const mint a)
    {
        if ((x += mod - a.x) >= mod)
            x -= mod;
        return *this;
    }
    mint &operator*=(const mint a)
    {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const { return mint(*this) += a; }
    mint operator-(const mint a) const { return mint(*this) -= a; }
    mint operator*(const mint a) const { return mint(*this) *= a; }
    mint pow(ll t) const
    {
        if (!t)
            return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1)
            a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2); }
    mint &operator/=(const mint a) { return *this *= a.inv(); }
    mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream &operator>>(istream &is, const mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

ll gcd(ll a, ll b)
{
    if (a == 0 || b == 0)
        return 1;
    if (a % b == 0)
        return (b);
    else
        return (gcd(b, a % b));
}

int main()
{
    int n;
    cin >> n;
    ll x, y, zero = 0;
    map<pair<ll, ll>, pi> iwashi; //回転して一致したら同じパラメータを持つのでmap
    for (int i = 0; i < n; i++)
    {
        cin >> x >> y;
        if (x == 0 && y == 0) //例外処理、(0,0)はどのイワシとも仲が悪い
        {
            zero++;
            continue;
        }
        ll g = gcd(x, y);
        x /= g;
        y /= g; //互いに素にする
        if (y < 0)
            x = -x, y = -y; //第3,4象限は180°回転
        else if (y == 0 && x < 0)
            x = -x, y = -y;
        bool rot90 = (x <= 0); //第2象限なら-90°回転
        if (rot90)
        {
            ll tmp = x;
            x = y, y = -tmp;
        }
        if (rot90)
            iwashi[{x, y}].first++;
        else
            iwashi[{x, y}].second++;
    }

    mint ans = 1;
    for (auto p : iwashi)
    {
        int s = p.second.first;
        int t = p.second.second;
        mint now = 1; //誰も選ばない
        now += mint(2).pow(s) - 1;
        now += mint(2).pow(t) - 1;
        ans *= now;
    }
    ans -= 1;
    ans += zero;
    cout << ans << endl;
}