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)

/*
confirm 0LL and 1LL
confirm cornercases such as 0
confirm times of cin < 10^6
*/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll, ll>;
using Pld = pair<ld, ld>;
using Vec = vector<ll>;
using VecP = vector<P>;
using VecB = vector<bool>;
using VecC = vector<char>;
using VecD = vector<ld>;
using VecS = vector<string>;
using Graph = vector<VecP>;
template <typename T>
using Vec1 = vector<T>;
template <typename T>
using Vec2 = vector<Vec1<T> >;
#define REP(i, m, n) for(ll i = (m); (i) < (n); ++(i))
#define REPN(i, m, n) for(ll i = (m); (i) <= (n); ++(i))
#define REPR(i, m, n) for(ll i = (m)-1; (i) >= (n); --(i))
#define REPNR(i, m, n) for(ll i = (m); (i) >= (n); --(i))
#define rep(i, n) REP(i, 0, n)
#define repn(i, n) REPN(i, 1, n)
#define repr(i, n) REPR(i, n, 0)
#define repnr(i, n) REPNR(i, n, 1)
#define all(s) (s).begin(), (s).end()
#define pb push_back
#define fs first
#define sc second
template <typename T>
bool chmax(T &a, const T b){if(a < b){a = b; return true;} return false;}
template <typename T>
bool chmin(T &a, const T b){if(a > b){a = b; return true;} return false;}
template <typename T>
ll pow2(const T n){return (1LL << n);}
template <typename T>
void cosp(const T n){cout << n << ' ';}
void co(void){cout << '\n';}
template <typename T>
void co(const T n){cout << n << '\n';}
template <typename T1, typename T2>
void co(pair<T1, T2> p){cout << p.fs << ' ' << p.sc << '\n';}
template <typename T>
void co(const Vec1<T> &v){for(T i : v) cosp(i); co();}
template <typename T>
void co(initializer_list<T> v){for(T i : v) cosp(i); co();}
template <typename T>
void ce(const T n){cerr << n << endl;}
void sonic(){ios::sync_with_stdio(false); cin.tie(0);}
void setp(const ll n){cout << fixed << setprecision(n);}
constexpr int INF = 1e9+1;
constexpr ll LINF = 1e18+1;
constexpr ll MOD = 1e9+7;
//constexpr ll MOD = 998244353;
constexpr ld EPS = 1e-11;
const ld PI = acos(-1);

ll gcd(ll a, ll b){
	if(!b) return a;
	return gcd(b, a%b);
}

ll PowMod(ll a, ll n){
	if(n < 0) return PowMod(PowMod(a, -n), MOD-2);
	if(n == 0) return 1;
	if(n == 1) return a;
	if(n%2 == 0) return PowMod(a*a%MOD, n/2);
	return a*PowMod(a*a%MOD, n/2)%MOD;
}

ll inv(ll a){
	return PowMod(a, MOD - 2);
}

int main(void){
	ll n;
	cin >> n;
	Vec a(n), b(n);
	rep(i, n) cin >> a[i] >> b[i];
	rep(i, n){
		ll g = gcd(a[i], b[i]);
		if(g == 0) continue;
		a[i] /= g;
		b[i] /= g;
	}

	map<P, ll> plus, minus;
	ll cntA = 0, cntB = 0, cntZ = 0;
	ll ans = 1;
	rep(i, n){
		if(a[i] == 0 && b[i] == 0){
			cntZ++;
			continue;
		}
		ll ng = 0;
		ll absA = abs(a[i]), absB = abs(b[i]);
		if(a[i] * b[i] > 0){
			ng += minus[{absB, absA}];
			plus[{absA, absB}]++;
		}else if(a[i] * b[i] < 0){
			ng += plus[{absB, absA}];
			minus[{absA, absB}]++;
		}else{
			if(a[i] == 0){
				ng += cntB;
				cntA++;
			}else{
				ng += cntA;
				cntB++;
			}
		}
		if(ng == 0) ans *= 2;
		else ans += ans * inv(PowMod(2, ng));
		ans %= MOD;
		cerr << ng << " " << ans << endl;
	}
	co((ans - 1 + cntZ) % MOD);

	return 0;
}