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: Accepted)

/*
author: madhav_1999 aka orthodoxparadox
17 May 2020 at 5:45 PM
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define dbl long double
#define int ll
#define ll long long
#define pii pair<int, int>
#define len(x) (int) x.size()
#define rev(a) reverse(all(a))
#define oiint ostream_iterator<int>(cout, " ") 
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define initialise(a, x) memset(a, x, sizeof(a))
#define onlyunique(v) v.erase(unique(all(v)), v.end()); //only for sorted vector
#define inf 2e18
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _CRT_SECURE_NO_WARNINGS
#ifdef MADHAV
#define dbg(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){cerr << name << " : " << arg1 << endl;}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');
	cerr.write(names, comma - names) << " : " << arg1 << "  ";
	__f(comma + 1, args...);}
#else
#define dbg(...) 42
#endif
template <typename X> ostream& operator << (ostream& x,const vector<X>& v){for(int i=0;i<(int)v.size();++i) x<<v[i]<<" ";return x;} 
template <typename X> ostream& operator << (ostream& x,const set<X>& v){for(auto it:v) x<<it<<" ";return x;} 
template <typename X> ostream& operator << (ostream& x,const multiset<X>& v){for(auto it:v) x<<it<<" ";return x;} 
template <typename X, typename Y> ostream& operator << (ostream& x, const pair<X,Y>& v) {x<<v.ff<<" "<<v.ss;return x;} 
template <typename T, typename S> ostream& operator << (ostream& os, const map<T, S>& v) { for (auto it : v) os << it.first << "=>" << it.second << endl; return os; }
typedef tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
struct pair_hash {inline std::size_t operator()(const std::pair<int, int> & v) const {return v.first*31+v.second;}};
const int MOD = 1e9 + 7;

int takemod(int a, int mod = MOD){a%=mod;if(a<0)a+=mod;return a;}

int fast_exp(int base, int expo, int mod = MOD) {
	int res=1;
	while(expo>0) {
		if(expo&1) res=(res*base)%mod;
		base=(base*base)%mod;
		expo>>=1;}
	return res;}

int modinv(int a, int mod = MOD){return takemod(fast_exp(takemod(a, mod), mod-2, mod), mod);}


void solve()
{
	int n;
	cin >> n;
	int a[n], b[n];
	int zcnta = 0, zcntb = 0, zcnt = 0;
	map<pii, pii> X;
	for(int i = 0; i < n; i++) {
		cin >> a[i] >> b[i];
		if(a[i] == 0 and b[i] == 0) {
			zcnt++;
			continue;
		}
		if(a[i] == 0) {
			zcnta++;
			continue;
		}
		if(b[i] == 0) {
			zcntb++;
			continue;
		}
		int x = __gcd(a[i], b[i]);
		a[i]/=x, b[i]/=x;
		if(b[i] < 0) {
			a[i] *= -1;
			b[i] *= -1;
		}
		if(a[i] < 0) {
			X[mp(b[i], -a[i])].ff++;
		}
		else {
			X[mp(a[i], b[i])].ss++;
		}
	}
	int ans = fast_exp(2, zcnta);
	ans = takemod(ans + fast_exp(2, zcntb));
	ans = takemod(ans - 1);
	for(auto r : X) {
		pii p = r.ss;
		dbg(r, p);
		int cur = fast_exp(2, p.ff);
		cur = takemod(cur + fast_exp(2, p.ss));
		cur = takemod(cur - 1);
		ans = takemod(ans * cur);
	}
	ans = takemod(ans + zcnt);
	cout << takemod(ans - 1);
}

signed main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	#ifdef MADHAV
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);
	#endif
	clock_t t1, t2;
	t1 = clock();

	int tt = 1;
	// cin >> tt;
	while(tt--)
	{
		solve();
	}
	#ifdef MADHAV
	t2 = clock(); cout << "\n\ntime taken: " << (t2-t1)/(dbl)CLOCKS_PER_SEC;
	#endif
	return 0;
}