#include <bits/stdc++.h>
#define rep(a,n) for (ll a = 0; a < (n); ++a)
using namespace std;
typedef long long ll;
using ll = long long;
typedef pair<ll,ll> P;
typedef pair<P,ll> PP;
using Graph = vector<vector<ll> >;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const ll INF = 1e18;

// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 1000000007;
//const int mod = 998244353;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  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 { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= 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 { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

long long modpow(long long a, long long n) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

// combination mod prime
// mintが必要
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
};


//input
ll n;


int main(){
    cin >> n;
    ll cr=0;
    ll cl=0;
    map<P,ll>mp1;//積が正
    map<P,ll>mp2;//積が負
    ll cnt = 0;
    rep(i,n){
        ll a,b;
        cin >> a >> b;
        if(a==0&&b==0){
            n--;
            cnt++;
            continue;
        }
        if(a==0){
            cr++;
            continue;
        }
        if(b==0){
            cl++;
            continue;
        }
        bool flag=false;
        if(a<0||b<0)flag=true;//負の数を持つかどうか
        a=abs(a);
        b=abs(b);
        ll g = __gcd(a,b);
        a /= g;
        b /= g;
        if(flag)b = -b;
        if(flag)mp2[{a,b}]++;
        else mp1[{a,b}]++;
    }
    mint ans = 1;
    ans *= (modpow(2,cr)+modpow(2,cl)-1);
    n -= (cr+cl);
    for(auto it=mp1.begin();it!=mp1.end();it++){
        ll now = it->second;
        ll opo = mp2[{it->first.second,-(it->first.first)}];
        n -= (now+opo);
        ans *= (modpow(2,now)+modpow(2,opo)-1);
    }
    ans *= modpow(2,n);
    ans += cnt;//1匹だけなら大丈夫
    cout << ans-1 << endl;

    return 0;
}