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: Runtime Error)

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using namespace boost::multiprecision;

# define REP(i,n) for (int i=0;i<(n);++i)
# define rep(i,a,b) for(int i=a;i<(b);++i)
# define all(v) v.begin(),v.end()
# define showVector(v) REP(i,v.size()){cout << (v[i]) << " ";} cout << endl;
template<class T> inline bool chmin(T &a, T b){ if(a > b) { a = b; return true;} return false;}
template<class T> inline bool chmax(T &a, T b){ if(a < b) { a = b; return true;} return false;}
typedef long long int ll;
typedef pair<ll,ll> P_ii;
typedef pair<double,double> P_dd;

template<class T>
using MaxHeap = std::priority_queue<T>;

template<class T>
using MinHeap = std::priority_queue<T, std::vector<T>, std::greater<T>>;

template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}

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

int pop_count(int n){
  return __builtin_popcount(n);
}

int pop_count(ll n){
  return __builtin_popcountll(n);
}

ll gcd(ll a, ll b) {
    if(a < b) swap(a,b);
    
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b){
    ll g = gcd(a,b);
    return (a/g)*b;
}

// 素数判定 O(√n)
bool is_prime(ll n){
    if(n == 1) return false;
    for(ll i = 2; i * i <= n; i++){
        if(n % i == 0) return false;
    }
    return true;
}

// 約数列挙 O(√n)
vector<ll> divisor(ll n){
    vector<ll> res;
    for(ll i = 1; i * i <= n; i++){
        if(n % i == 0){
            res.push_back(i);
            if(i != n / i) res.push_back(n / i);
        }
    }
    return res;
}

template<typename T>
map<T, ll> prime_factorize(T x){
  map<T, ll> res;
  if(x == 0) return res;
  while(x%2==0){
    x/=2;
    res[2]++;
  }
  for(ll i=3;i*i<=x;i+=2){
    while(x%i==0){
      x/=i;
      res[i]++;
    }
  }
  if(x!=1) res[x]++;
  return res;
}

// エラトステネスの篩
// return lp[i] := iの最小素因数(lp[i] == iならば素数)
vector<int> SieveOfEratosthenes(int N = 10000000){
  vector<int> lp(N + 1, 0);
  vector<int> pr;
  for (int i=2; i<=N; ++i) {
      if (lp[i] == 0) {
          lp[i] = i;
          pr.push_back(i);
      }
      for (int j=0; j<(int)pr.size() && pr[j]<=lp[i] && i*pr[j]<=N; ++j)
          lp[i * pr[j]] = pr[j];
  }
  return lp;
}


// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x(x%mod){}
  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 {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=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 {
    mint res(*this);
    return res/=a;
  }
};

const int MOD = 1000000007;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;

ll intceil(ll a, ll b){
    if(a < 0){
      a *= -1;
      b *= -1;
    }
    if(b >= 0) return (a + b - 1 ) / b;
    return a / b;
}

ll intfloor(ll a, ll b){
    if(a < 0){
      a *= -1;
      b *= -1;
    }
    if(b >= 0) return a / b;
    return (a - b + 1) / b;
}

void addM(ll &a, ll b) {
    a += b;
    if (a >= MOD) a -= MOD;
}

void mulM(ll &a, ll b) {
    a = ((a%MOD)*(b%MOD))%MOD ;
}

ll powM(ll a,ll b) {
    ll ret = 1;
    ll tmp = a;
    while(b>0) {
        if((b&1)==1) ret = (ret * tmp) % MOD;
        tmp = (tmp * tmp) % MOD;
        b = b >> 1;
    }
    return ret;
}

// mod. m での a の逆元 a^{-1} を計算する
ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

// ラングレンス圧縮
vector<pair<char, int>> rang_com(string s){
  vector<pair<char, int>> ret;
  string t = s;
  t.erase(unique(all(t)), t.end());
  int now = 0;
  int pre = 0;
  for(auto ct : t){
    while(now < s.size() && s[now] == ct) now++;
    if(ret.size() == 0){
      ret.push_back({ct, now});
    } else {
      ret.push_back({ct, now - pre});
    }
    pre = now;
  }
  return ret;
}

ll mypow(ll a, ll t) {
  if (!t) return 1;
  ll ret = mypow(a, t>>1);
  ret *= ret;
  if (t&1) ret *= a;
  return ret;
}

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

  int n;
  cin >> n;

  mint cnt = 0;
  map<pair<int, int>, pair<int, int>> mp;
  REP(i, n){
    int a, b;
    cin >> a >> b;
    if(a == 0 && b == 0){
      cnt += 1;
    } else {
      int g = __gcd(a, b);
      a /= g;
      b /= g;
      if(b < 0){
        // bを必ず正にする
        a = -a;
        b = -b;
      }
      if(a < 0){
        mp[{b, -a}].second++;
      } else {
        mp[{a, b}].first++;
      }
    }
  }

  mint ans = 1;
  for(auto m : mp){
    ans *= (powM(2LL, m.second.first) + powM(2LL, m.second.second) - 1);
  }
  ans += cnt - 1;
  cout << ans.x << endl;

  return 0;
}