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;

#define int long long
 
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

#define YES puts("YES")
#define Yes puts("Yes")
#define NO  puts("NO")
#define No  puts("No")
#define ALL(v) (v).begin(), (v).end()
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
 
#if 1
//-------------
// DUMPマクロ
// https://www.creativ.xyz/dump-cpp-652/

#define repi(itr, ds) for (auto itr = ds.begin(); itr != ds.end(); itr++)

// vector
template <typename T>
istream& operator>>(istream& is, vector<T>& vec) {
    for (T& x : vec) is >> x;
    return is;
}
// pair
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}
// vector
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
    os << "{";
    for (int i = 0; i < vec.size(); i++) {
        os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
    }
    os << "}";
    return os;
}
// map
template <typename T, typename U>
ostream& operator<<(ostream& os, map<T, U>& map_var) {
    os << "{";
    repi(itr, map_var) {
        os << *itr;
        itr++;
        if (itr != map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// set
template <typename T>
ostream& operator<<(ostream& os, set<T>& set_var) {
    os << "{";
    repi(itr, set_var) {
        os << *itr;
        itr++;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
 
#define DUMPOUT cerr
 
void dump_func() {
    DUMPOUT << endl;
}
template <class Head, class... Tail>
void dump_func(Head&& head, Tail&& ... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) > 0) {
        DUMPOUT << ", ";
    }
    dump_func(std::move(tail)...);
}
#ifdef DEBUG
#define DEB
#define dump(...)                                                              \
    DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                            \
            << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]"        \
            << endl                                                            \
            << "    ",                                                         \
        dump_func(__VA_ARGS__)
#else
#define DEB if (false)
#define dump(...)
#endif

// http://beet-aizu.hatenablog.com/entry/2018/04/08/145516
template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}

template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}

template<typename T,typename V>
typename enable_if<is_class<T>::value==0>::type
fill_v(T &t,const V &v){t=v;}

template<typename T,typename V>
typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t,const V &v){
  for(auto &e:t) fill_v(e,v);
}

#endif


//*
#define mod 1000000007 /*/
#define mod 998244353 //*/

typedef pair<int, int> P;
#define INF (1LL<<60)

template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};

using mint = Fp<mod>;
void solve(){
    int N;
    cin >> N;
    map<P,int> m;
    
    int cnt0 = 0;
    rep(i,N){
        int A, B;
        cin >> A >> B;
        int g = gcd(abs(A), abs(B));
        if(A==0 && B==0){
            cnt0++;
        }else{
            A/=g; B/=g;
            if(B<0){ A=-A; B=-B;}
            m[{A,B}]++;
        }
        dump(A,B);
    }
        
    mint ans = 1;
    for(auto i=m.begin(); i!=m.end(); i++){
        P p = i->first;
        int c = i->second;
        if(c==0) continue;
        
        P p2 = {-p.second, p.first};
        if(p2.second < 0) p2 = {-p2.first, -p2.second};
        int c2 = m[p2];
        
        m[p] = 0;
        m[p2] = 0;
        dump(p, c, c2);
        
        ans *= modpow(mint(2), c) + modpow(mint(2), c2)-1;
        
        
    }
    ans-=1;
    ans+=cnt0;
    
    cout << ans.val << endl;
}

signed main(){
    cout << fixed << setprecision(18);
    cerr << fixed << setprecision(18);
    solve();
}