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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

N = int(input())
data = [list(map(int,input().split())) for i in range(N)]
mod = 10**9+7
ans = 0

from math import gcd

A = []

for i in range(N):
    if data[i][0] == 0 and data[i][1] == 0:  #(0,0)の場合は以降取り除いて考える
        ans += 1
    else:                                 #各々の成分をgcdで割り、第1成分が非負になるように調整
        if data[i][0] < 0:
            data[i][0] *= -1
            data[i][1] *= -1
        g = gcd(data[i][0],data[i][1])
        data[i][0] //= g
        data[i][1] //= g
        if data[i][0] == 0:
            A.append((0,1))
        elif data[i][1] > 0:
            A.append((data[i][0],data[i][1]))
        elif data[i][1] == 0:
            A.append((1,0))
        else:
            A.append((data[i][0],data[i][1]))


from collections import Counter

C = Counter(A)
D = []             #仲が悪いイワシの数の組

for i,j in list(C.keys()):
    if j <= 0:
        continue
    else:
        if j == 0:
            if C[(0,1)] > 0:
                D.append((C[(1,0)],C[(0,1)]))
        else:
            if C[(j,-i)] > 0:
                D.append((C[(i,j)],C[(j,-i)]))

r = 0  #仲が悪くなり得るイワシの数
for i,j in D:
    r += i+j

a = 1  #仲が悪くなり得るイワシたちの選び方
for i,j in D:
    a = (a*(pow(2,i,mod)+pow(2,j,mod)-1)) % mod

a = (a*pow(2,len(A)-r,mod)) % mod  #仲が悪くなり得ないイワシたちの選び方

ans += a-1  #全部選ばないものを除く

print(ans)