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>
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
//#include "boost/multiprecision/cpp_int.hpp"
//typedef boost::multiprecision::cpp_int LL;
typedef long double dd;
#define i_7 (ll)(1E9+7)
//#define i_7 998244353
#define i_5 i_7-2
ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
typedef pair<dd,dd> d_d;
ll inf=(ll)1E16;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-9;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define fi first
#define se second
#define endl "\n"
#define SORT(v) sort(v.begin(),v.end())
#define ERASE(v) v.erase(unique(v.begin(),v.end()),v.end())
#define POSL(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define POSU(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

//////////////////////////


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

ll po(ll i,ll p){
    if(p==0)return 1;
    else{
        i=mod(i);
        if(p==1)return i;
        if(p%2==0)return po(mod(i*i),p/2);
        return mod(i*po(i,p-1));
    }
}
ll bunbo(ll n){
    return po(n,i_5);
}
#define N 200005
ll n;
ll a[N],b[N];
void guc(){
    ll cnt=0;
    rep(i,1,(1<<n)-1){
        vector<l_l>v;
        rep(j,0,n-1){
            if((i>>j)&1)v.pb({a[j],b[j]});
        }
        ll vs=v.size();
        bool f=true;
        rep(j,0,vs-1){
            rep(k,j+1,vs-1){
                if(v[j].fi*v[k].fi+v[j].se*v[k].se==0)f=false;
            }
        }
        if(f)cnt++;
    }
    cout<<cnt<<endl;
}

int main(){fastio
    
    cin>>n;
    rep(i,0,n-1)cin>>a[i]>>b[i];
    //guc();
    bool f=false;
    ll A=0,B=0;
    vector<l_l>v;
    rep(i,0,n-1){
        if(a[i]==0&&b[i]==0){
            f=true;
            continue;
        }
        if(a[i]==0){
            A++;
        }else if(b[i]==0){
            B++;
        }else{
            if(b[i]<0){
                b[i]*=-1;
                a[i]*=-1;
            }
            ll g=gcd(abs(a[i]),abs(b[i]));
            a[i]/=g;
            b[i]/=g;
            v.pb({a[i],b[i]});
        }
    }
    sort(v.begin(),v.end());
    ll ans=po(2,A)+po(2,B)-1;
    ans=mod(ans);
    ll vs=v.size();
    
    map<l_l,ll>mp;
    rep(i,0,vs-1){
        ll c=1;
        while(i<=vs-2){
            if(v[i]==v[i+1]){
                i++;c++;
            }else break;
        }
        mp[v[i]]=c;
    }
    
    set<l_l>used;
    for(auto X:mp){
        l_l x=X.fi;
        ll cnt=X.se;
        if(cnt==0)continue;
        if(used.find(x)!=used.end())continue;
        l_l ser={-x.se,x.fi};
        if(ser.se<0){
            ser.fi*=-1;
            ser.se*=-1;
        }
        
        if(mp.find(ser)==mp.end()){
            ans=mod(ans*po(2,cnt));
        }else{
            ll op=mp[ser];
            ll ad=po(2,cnt)+po(2,op)-1;
            ad=mod(ad);
            ans=mod(ans*ad);
            used.insert(ser);
        }
        
    }
    if(f)ans++;
    cout<<mod(ans-1)<<endl;
    
    return 0;
}
/*
 5
 0 0
 1 0
 0 12
 1 21
 -1 21
 */