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 Ruby (2.7.1) (judge verdict: Wrong Answer)

module ModInt
  MOD = 10 ** 9 + 7

  def to_mod(a); a % MOD; end
  def add(a, b); (a + b) % MOD; end
  def sub(a, b); (a - b) % MOD; end
  def mul(a, b); (a * b) % MOD; end
  def inv(a); a.pow(MOD - 2, MOD); end
  def div(a, b); mul(a, inv(b)); end
end

class FactTable
  include ModInt

  def initialize(n)
    @fact = [1] * (n + 1)
    @finv = [1] * (n + 1)
    2.upto(n) { |i| @fact[i] = mul(@fact[i - 1], i) }
    @finv[n] = inv(@fact[n])
    n.downto(3) { |i| @finv[i - 1] = mul(@finv[i], i) }
  end

  def nCk(n, m)
    mul(mul(@fact[n], @finv[m]), @finv[n - m])
  end
end

SNG = 10**20r # 特異点
INF = 10**19r

n = gets.to_i
iwashi = n.times.map do
  a,b = gets.split.map(&:to_r)
  if b == 0
    if a == 0
      SNG
    else a > 0
      INF
    end
  else
    Rational(a/b)
  end
end

def inv(a)
  if a == SNG
    SNG
  elsif a == 0r
    INF
  else
    -1r/a
  end
end

cnt = iwashi.tally
# bad = iwashi.map{|v|inv(v)}
seen = Hash.new(-1)

MOD = 1000000007
include ModInt

ans = 1
cnt.each do |key,num|
  next if seen[key] == 1
  next if key == SNG
  seen[key] = 1

  # 嫌いな相手はいるか？
  bad = inv(key)
  if cnt.has_key?(bad)
    seen[bad] = 1
    a = add(2.pow(num, MOD), 2.pow(cnt[bad], MOD))
    a = sub(a, 1)
    ans = mul(ans, a)
  else
    a = 2.pow(num, MOD)
    ans = mul(ans, a)
  end
end

# 特異点イワシ
if cnt.has_key?(SNG)
  ans += cnt[SNG]
end

# 何も選ばない場合を引く
ans = sub(ans,1)

puts ans