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)

import sys, bisect, math, itertools, string, queue, copy
# import numpy as np
# import scipy
from collections import Counter,defaultdict,deque
from itertools import permutations, combinations, combinations_with_replacement
from heapq import heappop, heappush
from fractions import gcd
# input = sys.stdin.readline
sys.setrecursionlimit(10**8)
mod = 10**9+7
def inp(): return int(input())
def inpm(): return map(int,input().split())
def inpl(): return list(map(int, input().split()))
def inpls(): return list(input().split())
def inplm(n): return list(int(input()) for _ in range(n))
def inplL(n): return [list(input()) for _ in range(n)]
def inplT(n): return [tuple(input()) for _ in range(n)]
def inpll(n): return [list(map(int, input().split())) for _ in range(n)]
def inplls(n): return sorted([list(map(int, input().split())) for _ in range(n)])

def main():
    n = inp()
    dic = defaultdict(int)
    dic1 = defaultdict(int)
    for _ in range(n):
        a,b = inpm()
        if a != 0 and b == 0:
            a = 1
        elif a == 0 and b != 0:
            b = 1
        else:
            if not (a==0 and b==0):
                m = gcd(a,b)
                a //= m
                b //= m
        if a < 0 and b < 0:
            a = -a
            b = -b
        if a < 0 and b > 0:
            a = -a
            b = -b
        index = ' '.join([str(a),str(b)])
        dic[index] += 1
        dic1[index] += 1
    ans = 1
    dic2 = defaultdict(int)
    for e in dic:
        a,b = map(int,e.split())
        if a == 0 or b == 0:
            continue
        flag = 0
        if a > 0 and b > 0:
            rev = ' '.join([str(b),str(-a)])
        else:
            rev = ' '.join([str(-b),str(a)])
        if dic1[rev] > 0 and dic2[rev] == 0:
            dic2[rev] = 1
            dic2[e] = 1
            flag = 1
        if flag:
            continue
        n1 = pow(2,dic[e],mod)
        n2 = pow(2,dic1[rev],mod)
        if n1 == 0:
            ans *= n2
        elif n2 == 0:
            ans *= n1
        else:
            ans *= (n1+n2-1)
        ans %= mod
    num = 1
    if dic['0 0']:
        num *= (dic['0 0']+1)
        num %= mod
    if dic['1 0'] and dic['0 1']:
        num += (pow(2,dic['1 0'],mod)+pow(2,dic['0 1'],mod)-1)
        num %= mod
    elif dic['1 0']:
        num += pow(2,dic['1 0'],mod)
        num %= mod
    elif dic['0 1']:
        num += pow(2,dic['0 1'],mod)
        num %= mod
    ans *= num
    ans %= mod
    print(ans-1)
    
if __name__ == "__main__":
    main()