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: Accepted)

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
//typedef vector<vector<ll>> Graph;

const ll mod = 1e9 + 7;
//const ll mod = 998244353;

#define REP(i,n) for(ll i=0;i<(ll)n;i++)
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define spa << " " <<
#define fi first
#define se second

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; }

template<class S, class T> ostream& operator << (ostream& os, const pair<S, T> v){
os << "(" << v.first << ", " << v.second << ")"; return os;
}
template<class T> ostream& operator << (ostream& os, const vector<T> v){
for(int i = 0; i < (int)v.size(); i++){if(i > 0){os << " ";} os << v[i];} return os;
}
template<class T> ostream& operator << (ostream& os, const vector<vector<T>> v){
for(int i = 0; i < (int)v.size(); i++){if(i > 0){os << endl;} os << v[i];} return os;
}

template<typename T> void debug(vector<vector<T>>&v,ll h,ll w){for(ll i=0;i<h;i++)
{cerr<<v[i][0];for(ll j=1;j<w;j++)cerr spa v[i][j];cerr<<endl;}};
template<typename T> void debug(vector<T>&v,ll n){if(n!=0)cerr<<v[0];
for(ll i=1;i<n;i++)cerr spa v[i];
cerr<<endl;};

string num2bit(ll num, ll len){
    string bit = "";
    REP(i, len){
    bit += char('0'+(num>>i & 1));
    }
    return bit;
}

template <std::uint_fast64_t Modulus> class modint {
  // long long から modint を作るときは必ず正の数にしてからコンストラクタに入れること！ 
  // そうしないとバグります
  using u64 = std::uint_fast64_t;

public:
  u64 a;

  constexpr modint(const u64 x = 0) noexcept : a(x % Modulus) {}
  constexpr u64 &value() noexcept { return a; }
  constexpr const u64 &value() const noexcept { return a; }
  constexpr modint operator+(const modint rhs) const noexcept {
    return modint(*this) += rhs;
  }
  constexpr modint operator-(const modint rhs) const noexcept {
    return modint(*this) -= rhs;
  }
  constexpr modint operator*(const modint rhs) const noexcept {
    return modint(*this) *= rhs;
  }
  constexpr modint operator/(const modint rhs) const noexcept {
    return modint(*this) /= rhs;
  }
  constexpr modint &operator+=(const modint rhs) noexcept {
    a += rhs.a;
    if (a >= Modulus) {
      a -= Modulus;
    }
    return *this;
  }
  constexpr modint &operator-=(const modint rhs) noexcept {
    if (a < rhs.a) {
      a += Modulus;
    }
    a -= rhs.a;
    return *this;
  }
  constexpr modint &operator*=(const modint rhs) noexcept {
    a = a * rhs.a % Modulus;
    return *this;
  }
  constexpr modint &operator/=(modint rhs) noexcept {
    u64 exp = Modulus - 2;
    while (exp) {
      if (exp % 2) {
        *this *= rhs;
      }
      rhs *= rhs;
      exp /= 2;
    }
    return *this;
  }
};
using mint = modint<mod>;

template <class T, class U> constexpr T power(T x, U exp) {
  T ret = static_cast<T>(1);
  while (exp) {
    if (exp % static_cast<U>(2) == static_cast<U>(1))
      ret *= x;
    exp /= static_cast<U>(2);
    x *= x;
  }
  return ::std::move(ret);
}

ll Euclid(ll A, ll B){
    if(A<B) swap(A, B);
    ll C = B;
    while(C != 0){
        C = A%B;
        A = B;
        B = C;
    }
    return A;
}

// コーナー A=0 and B=0 の場合，A = 0 かつ B < 0 の場合

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N;
    cin >> N;

    map<pll, ll> cnt;
    ll z = 0;
    REP(i, N){
        ll A, B;
        cin >> A >> B;
        if(A<0){
            A *= -1;
            B *= -1;
        }else if(A==0 and B < 0){
            B *= -1;
        }else if(A==0 and B==0){
            z++;
            continue;
        }
        ll G = Euclid(abs(A), abs(B));
        if(G != 0){
            A /= G;
            B /= G;
        }
        pll p;
        p.first = A;
        p.second = B;
        cnt[p]++;
    }

    N -= z;

    mint res = 1;

    const ll INF = (1ll<<60);
    vector<pll> kumi;
    ll single = 0;
    for(auto x: cnt){
        ll c1 = x.second;
        if(c1 == -INF) continue;
        ll a = x.first.first;
        ll b = x.first.second;
        ll c = -b;
        ll d = a;
        if(c<0){
            c *= -1;
            d *= -1;
        }else if(c==0 and d<0){
            d *= -1;
        }
        pll q = {c, d};

        //ll c2 = cnt[q];
        if(cnt.find(q) != cnt.end()){
            ll c2 = cnt[q];
            kumi.push_back({c1, c2});
            cnt[q] = -INF;
        }else{
            single += c1;
        }

        //dump(x.first)
        //dump(q)
        //dump(c1)
    }

    //cout << kumi << endl;
    //dump(single)

    for(auto x: kumi){
        mint X1 = power(mint(2), x.first);
        X1 -= 1;
        mint X2 = power(mint(2), x.second);
        X2 -= 1;
        res *= (X1+X2+1);
    }

    res *= power(mint(2), single);
    res -= 1;

    res += z; //?
    cout << res.value() << endl;

    return 0;
}