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)

# E - ∙ (Bullet)
import sys
from collections import defaultdict
from math import gcd

readline = sys.stdin.readline
MOD = 1000000007
N = int(readline())

# イワシの傾きをA/Bとする。
# A/Bは予めgcdで割って約分し、同じ傾きのものを揃えておく。
# このとき、仲の悪い組合せは(A, B)に対して(B, -A)となるもの。 
# 前者をS、後者をTとして考える。
d_S = defaultdict(int)
d_T = defaultdict(int)

'''
def gcd(x, y):
    if x == 0:
        return y
    else:
        return gcd(y%x, x)
'''

pow2_mod = [1] + [0] * (N + 10)
for i in range(1, N + 10):
    pow2_mod[i] = pow2_mod[i-1] * 2 % MOD

zeros = 0
for i in range(N):
    A, B = map(int, readline().split())
    # (0, 0)のイワシは他の全てのイワシと仲が悪いため、
    # 1匹のみでしか選ぶことができない。最後にこの数を足す。
    if A == 0 and B == 0:
        zeros += 1
    # (?, 0),(0, ?)のイワシは、?がどのような数字であろうとも
    # この組合せであれば仲が悪い。一律、(1,0)(0,-1)として考える。
    elif A == 0:
        d_S[(1, 0)] += 0
        d_T[(0, -1)] += 1
    elif B == 0:
        d_S[(1, 0)] += 1
        #d_T[(0, -1)] += 0
    else:
        g = gcd(A, B)
        A //= g
        B //= g
        # いずれもマイナスの場合、傾きはプラスになる
        if A < 0 and B < 0:
            A *= -1
            B *= -1
        if A * B > 0:
            d_S[(A, B)] += 1
            #d_T[(B, -A)] += 0
        # 片方がマイナスの場合、必ずAがマイナスになるよう揃える
        else:
            if A > B:
                A, B = B, A
            d_S[(-A, B)] += 0
            d_T[(B, A)] += 1

ans = 1
for (si, sj), s_cnt in d_S.items():
    t_cnt = d_T[(sj, -1*si)]
    tmp = 0
    tmp += pow2_mod[s_cnt] - 1 # Sから1匹以上選ぶ組合せ
    tmp += pow2_mod[t_cnt] - 1 # Tから1匹以上選ぶ組合せ
    tmp += 1 # どちらも選ばない組合せ
    ans = (ans * tmp)%MOD

ans -= 1 # 最終的に1匹も選ばない組合せ
ans += zeros # (0, 0)のイワシの数

print(ans%MOD)