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: Accepted)

MOD = 10 ** 9 + 7
def pow(n, k)
  k %= (MOD - 1)
  return 1 if k.zero?
  return pow(n, k - 1) * n % MOD if k.odd?
  return (pow(n, k / 2) ** 2) % MOD
end
def max(a,b); a > b ? a : b; end
  
N = gets.to_i
ab0 = 0
a0 = 0
b0 = 0
dp = Hash.new(0)

N.times.each do
  a,b = gets.split.map(&:to_i)
  if a.zero? && b.zero?
    # どの魚とも仲が悪い
    ab0 += 1
  elsif a.zero?
    # b == 0の魚と仲が悪い
    a0 += 1
  elsif b.zero?
    # a == 0 の魚と仲が悪い
    b0 += 1
  else
    # a/bでグループ分け
    # -b/aのグループと仲が悪い
    dp[a.to_r / b] += 1
  end
end

# 一旦 どの魚とも仲が悪いのは無視して、
# 仲が悪いグループのどちらかを選ぶ組み合わせを計算 2^a + 2^b - 1
# 仲が悪いグループがいない場合 2^a
ans = 1
used = {}

if a0.zero? || b0.zero?
  ans = ans * pow(2, max(a0, b0)) % MOD
else
  ans = ans * (pow(2, a0) + pow(2,b0) - 1) % MOD
end
dp.keys.each do |k|
  next if used[k]
  used[k] = true
  if dp[-1/k] > 0
    ans = ans * (pow(2, dp[k]) + pow(2, dp[-1/k]) - 1) % MOD
    used[-1/k] = true
  else
    ans = ans * pow(2, dp[k]) % MOD
  end
end
# 何も選ばない場合を除く + どの魚とも仲が悪いグループを足す
puts (ans - 1 + ab0 ) % MOD