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>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) begin(v),end(v)
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
using ll = long long;
using pii = pair<int, int>;
constexpr ll INF = 1ll<<30;
constexpr ll longINF = 1ll<<60;
constexpr ll MOD = 1000000007;
constexpr bool debug = 0;
//---------------------------------//

template<int M>
struct ModInt {
public:
	using value_type = long long;
	
	ModInt(value_type val = 0) : val(val < 0 ? (M - (-val % M)) % M : val % M) {}
	
	explicit operator bool() const noexcept { return val; }
	bool operator ==(const ModInt & rhs) const noexcept { return val == rhs.val; }
	bool operator !=(const ModInt & rhs) const noexcept { return !(*this == rhs); }
	ModInt operator +() const noexcept { return ModInt(*this); }
	ModInt operator -() const noexcept { return ModInt(0) -= *this; }
	ModInt operator +(const ModInt & rhs) const noexcept { return ModInt(*this) += rhs; }
	ModInt operator -(const ModInt & rhs) const noexcept { return ModInt(*this) -= rhs; }
	ModInt operator *(const ModInt & rhs) const noexcept { return ModInt(*this) *= rhs; }
	ModInt operator /(const ModInt & rhs) const noexcept { return ModInt(*this) /= rhs; }
	
	ModInt & operator +=(const ModInt & rhs) noexcept {
		val += rhs.val;
		if (val >= M) val -= M;
		return *this;
	}
	ModInt & operator -=(const ModInt & rhs) noexcept {
		if (val < rhs.val) val += M;
		val -= rhs.val;
		return *this;
	}
	ModInt & operator *=(const ModInt & rhs) noexcept {
		val = val * rhs.val % M;
		return *this;
	}
	ModInt & operator /=(const ModInt & rhs) noexcept {
		*this *= rhs.inverse();
		return *this;
	}
	
	ModInt pow(value_type n) const {
		ModInt res = 1, x = val;
		if (n < 0) { x = x.inverse(); n = -n; }
		while (n) { if (n & 1) res *= x; x *= x; n >>= 1; }
		return res;
	}
	
	ModInt inverse() const {
		long long a = val, a1 = 1, a2 = 0, b = M, b1 = 0, b2 = 1;
		while (b > 0) {
			value_type q = a / b, r = a % b;
			value_type nb1 = a1 - q * b1, nb2 = a2 - q * b2;
			a = b; b = r;
			a1 = b1; b1 = nb1;
			a2 = b2; b2 = nb2;
		}
		assert(a == 1);
		return a1;
	}
	
	const value_type & get() const noexcept { return val; }
	static decltype(M) get_mod() noexcept { return M; }
	
	friend std::ostream & operator <<(std::ostream & os, const ModInt & rhs) { return os << rhs.val; }
	friend std::istream & operator >>(std::istream & is, ModInt & rhs) {
		value_type x;
		is >> x;
		rhs = ModInt(x);
		return is;
	}
private:
	value_type val;
};
using mint = ModInt<MOD>;

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

int main() {
	int N;
	cin >> N;
	
	vector<ll> A(N), B(N), C(N);
	REP(i, N) {
		scanf("%lld %lld", &A[i], &B[i]);
		C[i] = (A[i] > 0) ^ (B[i] > 0);
		A[i] = abs(A[i]);
		B[i] = abs(B[i]);
		
		if (A[i] > 0 || B[i] > 0) {
			ll g = gcd(A[i], B[i]);
			A[i] /= g;
			B[i] /= g;
		}
	}
	
	int ab0 = 0;
	using pll = pair<ll, ll>;
	map<pll, int> mm;
	REP(i, N) {
		if (A[i] == 0 && B[i] == 0) ++ab0;
		else ++mm[pll((C[i] ? -1 : 1) * A[i], B[i])];
	}
	
	mint ans = 1;
	int sum = 0;
	for (const auto &cur : mm) {
		int a = abs(cur.fi.fi);
		int b = cur.fi.se;
		int c = cur.fi.fi <= 0;
		if (c) continue;
		auto it = mm.find(pll(-b, a));
		int cnt = it == mm.end() ? 0 : it->se;
		ans *= (mint(2).pow(cur.se) - 1) + (mint(2).pow(cnt) - 1) + 1;
		sum += cur.se + cnt;
	}
	ans *= mint(2).pow(N - sum);
	ans = ans - 1;
	cout << ans << endl;
	return 0;
}