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>
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
#define lli long long int
#define uli unsigned long long int
#define INF 9999999999
#define rep(i,m,n) for(lli i = m;i < n;i++)
#define rrep(i,m,n) for(lli i=m-1;i>=n;i--)
#define pb(n) push_back(n)
#define UE(N) N.erase(unique(N.begin(),N.end()),N.end());
#define Sort(n) sort(n.begin(), n.end())
#define Rev(n) reverse(n.begin(),n.end())
#define Out(S) cout << S << endl
#define NeOut(S) cout << S
#define HpOut(S) cout << setprecision(25) << S << endl
#define Vec(K,L,N,S) vector<L> K(N,S)
#define DV(K,L,N,M,S) vector<vector<L>> K(N,vector<L>(M,S))
#define TV(K,L,N,M,R,S) vector<vector<vector<L>>> K(N,vector<vector<L>>(M,vector<L>(R,S)))
#define pint pair<lli,lli>
#define paf(L,R) pair<L,R>
#define mod 1000000007
#define MAX 10000000
#define ALL(a)  a.begin(),a.end()
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
lli repow(lli n,lli k,lli m){
  if(k==0)return 1;
  else if(k%2==1)return repow(n,k-1,m)*n%m;
  else{
    lli temp=repow(n,k/2,m);
    return temp*temp%m;
  }
}
lli gcd(lli num1, lli num2){
    return num2 != 0 ? gcd(num2, num1 % num2) : num1;
}
int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  lli A,B,C,D,E,F,N,M,K,L,X,Y,Z,H,W,sum=0,num=0,flag=0;string S,T;
  cin >> N;
  map<pint,int>mp1;
  map<pint,int>mp2;
  rep(i,0,N){
    cin >> A >> B;
    C=gcd(abs(A),abs(B));
    if(A==B&&A==0)flag++;
    else if(B==0)mp1[pint(1,0)]++;
    else if(A==0)mp2[pint(1,0)]++;
    else if((double)A/B>=0){
      A/=C;
      B/=C;
      if(A<0&&B<0){
        A=-A;
        B=-B;
      }
      mp1[pint(A,B)]++;
    }
    else{
      A/=C;
      B/=C;
      if(A<0){
        A=-A;
      }
      else B=-B;
      mp2[pint(B,A)]++;
    }
  }
  num=1;
  set<pint>st;
  for(auto v:mp1)st.insert(v.first);
  for(auto v:mp2)st.insert(v.first);
  for(auto v:st){
    if(mp1.find(v)!=mp1.end()&&mp2.find(v)!=mp2.end()){
      num*=(repow(2,mp1[v],mod)+repow(2,mp2[v],mod)-1)%mod;
    }
    else if(mp1.find(v)!=mp1.end()){
      num*=(mp1[v]+1);
    }
    else{
      num*=(mp2[v]+1);
    }
    num%=mod;
    Out(num);
  }
  Out((num+mod+flag-1)%mod);
}