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<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
#include<unordered_map>
#include<stack>
#include<string>
#include<iomanip>
#include<algorithm>
#include<functional>
#include<cstring>
#include<utility>
#include<math.h>
using namespace std;



/**** Type Define ****/

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> Q;

/**** Const List   ****/
const double PI=3.14159265358979323846;
const ll INF = 1LL << 60;
const ll mod = 1000000007;
const ll dx[4] = {1, 0, -1, 0};
const ll dy[4] = {0, -1, 0, 1};
const ll NCK_MAX = 510000;
/**** General Functions ****/
ll ketawa(ll n){
  ll a=0;
  while(n!=0){
    a+=n%10;
    n/=10;
  }
  return a;
}
ll RepeatSquaring(ll N, ll P, ll M){//pow->double
    if(P==0) return 1;
    if(P%2==0){
        ll t = RepeatSquaring(N, P/2, M);
        return (t% M)*(t % M)%M;
    }
    return (N * RepeatSquaring(N, P-1, M))%M;
}
ll RS(ll N, ll P){//modがだるいときにつかう
    if(P==0) return 1;
    if(P%2==0){
        ll t = RS(N, P/2);
        return t*t;
    }
    return (N * RS(N, P-1));
}
map< ll, int > prime_factor(ll n) {
  map< ll, int > ret;
  for(ll i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}
bool IsPrime(ll a){//order root a
  if(a==1)return false;
  for(int i=2;i*i<=a;i++){
    if(a%i==0&&a!=i){
      return false;
    }
  }
  return true;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll lcm(ll a, ll b){
    return a  / gcd(a, b)*b;
}

ll extgcd(ll a, ll b, ll& x, ll& y) {
  if (b == 0) {
    x = 1, y = 0; return a;
  }
  ll q = a/b, g = extgcd(b, a - q*b, x, y);
  ll z = x - q * y;
  x = y;
  y = z;
  return g;
}

ll invmod (ll a, ll m) { // a^-1 mod m
  ll x, y;
  extgcd(a, m, x, y);
  x %= m;
  if (x < 0) x += m;
  return x;
}

ll *fac, *finv, *inv;

void nCk_init(ll mod) {
  fac = new ll[NCK_MAX];
  finv = new ll[NCK_MAX];
  inv = new ll[NCK_MAX];
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (ll i = 2; i < NCK_MAX; i++) {
    fac[i] = fac[i-1] * i % mod;
    inv[i] = mod - inv[mod%i] * (mod / i) % mod;
    finv[i] = finv[i-1] * inv[i] % mod;
  }
}

ll nCk(ll n, ll k, ll mod) {
  if (fac == NULL) nCk_init(mod);
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n-k] % mod) % mod;
}

ll lmin(ll a, ll b) { return a > b ? b : a; };

ll lmax(ll a, ll b) { return a > b ? a : b; };

ll lsum(ll a, ll b) { return a + b; };

void warshall_floyd(int n) {
  for (int k = 0; k < n; k++){       // 経由する頂点
    for (int i = 0; i < n; i++) {    // 始点
      for (int j = 0; j < n; j++) {  // 終点
        //d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
      }
    }
  }
}

// 汎用的な二分探索のテンプレ
/*
int binary_search(int key) {
    ll ng = -1; //絶対falseの値、なければ最小値-1
    ll ok = (int)a.size(); // 絶対trueの値　なければ最大値+1

    // ok と ng のどちらが大きいかわからないことを考慮
    while (abs(ok - ng) > 1) {
        ll mid = (ok + ng) / 2;

        if (isOK(mid, key)) ok = mid;
        else ng = mid;
    }
    return ok;
}
*/
/**** Zip ****/

template <typename T>
class Zip {
  vector<T> d;
  bool flag;
public:
  Zip() {
    flag = false;
  }
  void add(T x) {
    d.push_back(x);
    flag = true;
  }
  ll getNum(T x) { // T need to have operator < !!
    if (flag) {
      sort(d.begin(), d.end());
      d.erase(unique(d.begin(), d.end()), d.end());
      flag = false;
    }
    return lower_bound(d.begin(), d.end(), x) - d.begin();
  }
  ll size() {
    if (flag) {
      sort(d.begin(), d.end());
      d.erase(unique(d.begin(), d.end()), d.end());
      flag = false;
    }
    return (ll)d.size();
  }
};

/**** Union Find ****/

class UnionFind {
  vector<ll> par, rank; // par > 0: number, par < 0: -par
public:
  void init(ll n) {
    par.resize(n, 1); rank.resize(n, 0);
  }
  ll getSize(ll x) {
    return par[find(x)];
  }
  ll find(ll x) {
    if (par[x] > 0) return x;
    return -(par[x] = -find(-par[x]));
  }
  void merge(ll x, ll y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (rank[x] < rank[y]) {
      par[y] += par[x];
      par[x] = -y;
    } else {
      par[x] += par[y];
      par[y] = -x;
      if (rank[x] == rank[y]) rank[x]++;
    }
  }
  bool isSame(ll x, ll y) {
    return find(x) == find(y);
  }
};

/**** Segment Tree ****/
class SegmentTree {
public:
  vector<pair<double, double>> node;//node[0]は使用しない
  ll n;//データの個数(nodeの最下層には何個並んでいるか)
  pair<double, double> initial_value;//初期値
public:
  void Init(ll n_,pair<double, double> initial_value_){
    n=1;
    while(n<n_)n*=2;
    node.resize(2*n);
    for(ll i=0;i<2*n;i++){
      node[i]=initial_value_;
    }
    initial_value=initial_value_;
  }
  void Update(ll k,pair<double, double> a){
    //node[k]をaにする
    //それに従って先祖も変わっていく
    k+=n;
    node[k]=a;
    while(k>1){
      k=k/2;
      node[k]=pair<double, double>(node[k*2].first*node[k*2+1].first,node[k*2].second*node[k*2+1].first+node[k*2+1].second);
    }
  }
  /*void Watch(){
    for(ll i=0;i<2*n;i++){
      cout<<node[i]<<endl;
    }
  }*/
  double Query(){//[a,b)
    return node[1].first+node[1].second;
  }
};


/**** LIS ****/
ll lis(ll* a, ll n, ll* dp) {
  fill(dp, dp + n, INF);//INFを代入
  for (ll i = 0; i < n; i++) *lower_bound(dp, dp + n, a[i]) = a[i];
  return (ll)(lower_bound(dp, dp + n, INF) - dp);
}


/**** main function ****/
ll n,m,k;
ll ans;
string s;
vector<ll> t,a;
ll kakutei[100001];
map<P,ll> mp;
int main(){
  cin>>n;
  for(ll i=0;i<n;i++){
    ll aa,bb;
    cin>>aa>>bb;
    if(aa==0&&bb==0){
      mp[P(0,0)]++;
    }else if(aa==0&&bb!=0){
      mp[P(0,1)]++;
    }else if(aa!=0&&bb==0){
      mp[P(1,0)]++;
    }else{
      ll mainasunanko=0;
      if(aa<0){
        mainasunanko++;
        aa*=-1;
      }
      if(bb<0){
        mainasunanko++;
        bb*=-1;
      }
      ll gcdd=gcd(aa,bb);
      aa/=gcdd;
      bb/=gcdd;
      if(mainasunanko%2==0){
        mp[P(aa,bb)]++;
      }else{
        mp[P(-1*aa,bb)]++;
      }
    }
  }
  ll zerozero=0;
  ans=1;
  for(auto itr=mp.begin();itr!=mp.end();itr++){
    ll aa=itr->first.first;
    ll bb=itr->first.second;
    ll value=itr->second;
    ll toori;
    if(value==-1)continue;
    if(aa==0&&bb==0){
      zerozero++;
      toori=1;
    }else if(aa==0&&bb!=0){
      toori=RepeatSquaring(2,value,mod)+RepeatSquaring(2,mp[P(1,0)],mod)-1;
      toori%=mod;
      mp[P(1,0)]=-1;
      mp[P(0,1)]=-1;
    }else if(aa!=0&&bb==0){
      toori=RepeatSquaring(2,value,mod)+RepeatSquaring(2,mp[P(1,0)],mod)-1;
      toori%=mod;
      mp[P(1,0)]=-1;
      mp[P(0,1)]=-1;
    }else{
      if(aa>0){
        toori=RepeatSquaring(2,value,mod)+RepeatSquaring(2,mp[P(-1*bb,aa)],mod)-1;
        toori%=mod;
        mp[P(aa,bb)]=-1;
        mp[P(-1*bb,aa)]=-1;
      }else{
        toori=RepeatSquaring(2,value,mod)+RepeatSquaring(2,mp[P(bb,aa*-1)],mod)-1;
        toori%=mod;
        mp[P(aa,bb)]=-1;
        mp[P(bb,aa*-1)]=-1;
      }
    }

    ans*=toori;
    ans%=mod;
  }
  cout<<ans+zerozero-1<<endl;
}