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 Python (3.8.2) (judge verdict: Wrong Answer)

def e_bullet(MOD=10**9 + 7):
    from math import gcd
    from collections import defaultdict
    N = int(input())

    zero_all = 0
    pair = defaultdict(int)
    for _ in range(N):
        a, b = [int(i) for i in input().split()]
        if a == 0 and b == 0:
            zero_all += 1
            continue

        # 既約分数にしておきたい
        g = gcd(a, b)
        pair[(a // g, b // g)] += 1  # 既約分数 a/b とみなす

    done = set()
    total = 1
    for (a, b), v in list(pair.items()):
        if (b, -a) in done or (-b, a) in done:
            continue  # a/b に対応するものがあるかはすでに調べた
        done.add((a, b))

        # a/b に対応する b/(-a) か (-b)/a に注目する
        w = pair[(b, -a)] + pair[(-b, a)]
        # それらを選ぶと仲が悪いので、a/b となるものだけ選ぶか b/(-a) か (-b)/a
        # となるものだけ選ぶか、どちらか一方だけ選択したときに仲が悪くならない
        # これまでの選び方に対して仲が悪くならない選び方を決めるので、total は積を取る
        # ただし、両方とも 1 つも選ばないということはできないため、-1 する
        total *= pow(2, v, MOD) + pow(2, w, MOD) - 1
        total %= MOD

    ans = (total + zero_all - 1) % MOD
    return ans

print(e_bullet())