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)

#pragma GCC optimize("O3")
#include<bits/stdc++.h> 
using namespace std;
using ll=long long;
typedef  unsigned long long int ull;
typedef pair<ll,ll> P;
template<class T> using V=vector<T>; 
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
const ll inf=(1e18);
//const ll mod=998244353;
const ll mod=1000000007;
ll GCD(ll a,ll b) {return b ? GCD(b,a%b):a;}
ll LCM(ll c,ll d){return c/GCD(c,d)*d;}
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
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; }
struct mint{
    using ull=unsigned long long int;
    ull v;
    mint(ll vv=0){s(vv%mod+mod);}
    mint& s(ull vv){
        v=vv<mod?vv:vv-mod;
        return *this;
    }
    //オーバーロード
    mint operator-()const{return mint()-*this;}//符号反転
    mint&operator+=(const mint&val){return s(v+val.v);}
    mint&operator-=(const mint&val){return s(v+mod-val.v);}
    mint&operator*=(const mint&val){
        v=ull(v)*val.v%mod;
        return *this;
    }
    mint&operator/=(const mint&val){return *this*=val.inv();}
    mint operator+(const mint&val){return mint(*this)+=val;}
    mint operator-(const mint&val){return mint(*this)-=val;}
    mint operator*(const mint&val){return mint(*this)*=val;}
    mint operator/(const mint&val){return mint(*this)/=val;}
    mint pow(ll n)const{
        mint res(1),x(*this);
        while(n){
            if(n&1)res*=x;
            x*=x;
            n>>=1;
        }
        return res;
    }
     mint inv()const{return pow(mod-2);}
    //拡張ユークリッドの互除法
    /* mint inv()const{
        int x,y;
        int g=extgcd(v,mod,x,y);
        assert(g==1);
        if(x<0)x+=mod;
        return mint(x);
    }*/
    friend ostream& operator<<(ostream&os,const mint&val){
        return os<<val.v;
    }//出力
    bool operator<(const mint&val)const{return v<val.v;}
    bool operator==(const mint&val)const{return v==val.v;}
    bool operator>(const mint&val)const{return v>val.v;}
};
map<P,P> mp;
int main(){
  ll n;
  cin>>n;
  ll zero=0;
  for(int i=0;i<n;i++){
    ll x,y;
    cin>>x>>y;
    if(x==0&&y==0){
      zero++;
      continue;
    }
    ll g=GCD(x,y);
    x/=g;y/=g;
    if(y<0){
      y=-y;
      x=-x;
    }
    if(y==0&&x<0){
      x=-x;
    }
    if(x<=0){
     mp[P(y,-x)].fi++;
    }else{
      mp[P(x,y)].se++;
    }
  }
  mint ans=1;
  for(auto p:mp){
      ll s=p.se.fi,t=p.se.se;
      mint now=1;
      now+=mint(2).pow(s)-mint(1);
      now+=mint(2).pow(t)-mint(1);
      ans*=now;
  }
  ans-=mint(1);
  ans+=mint(zero);
  cout<<ans<<"\n";
}