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 Lua (LuaJIT 2.1.0) (judge verdict: Wrong Answer)

local mod = 1000000007
local mfl = math.floor

local function bmul(x, y)
  local x1, y1 = mfl(x / 31623), mfl(y / 31623)
  local x0, y0 = x - x1 * 31623, y - y1 * 31623
  return (x1 * y1 * 14122 + (x1 * y0 + x0 * y1) * 31623 + x0 * y0) % mod
end

local function badd(x, y)
  return (x + y) % mod
end

local function bsub(x, y)
  return x < y and x - y + mod or x - y
end

local function modpow(src, pow)
  local res = 1
  while 0 < pow do
    if pow % 2 == 1 then
      res = bmul(res, src)
      pow = pow - 1
    end
    src = bmul(src, src)
    pow = mfl(pow / 2)
  end
  return res
end

local function modinv(src)
  return modpow(src, mod - 2)
end

local function getgcd(x, y)
  while 0LL < x do
    x, y = y % x, x
  end
  return y
end

local ffi = require("ffi")
local C = ffi.C
ffi.cdef[[
long long atoll(const char*);
]]

local function lltonumber(str)
  return C.atoll(str)
end

local n = io.read("*n", "*l")
local a, b = {}, {}
for i = 1, n do
  local s = io.read()
  local ai, bi = s:match("(%-?%d+) (%-?%d+)")
  a[i] = lltonumber(ai)
  b[i] = lltonumber(bi)
end

local x1, y1 = {}, {}
local x2, y2 = {}, {}
local xzero = 0
local yzero = 0

for i = 1, n do
  if b[i] < 0LL then
    a[i], b[i] = -a[i], -b[i]
  end
  if a[i] == 0LL and b[i] == 0LL then
  elseif a[i] == 0LL then
    xzero = xzero + 1
  elseif b[i] == 0LL then
    yzero = yzero + 1
  else
    if 0LL < a[i] then
      local xi = a[i]
      local yi = b[i]
      local gcd = getgcd(xi, yi)
      table.insert(x1, xi / gcd)
      table.insert(y1, yi / gcd)
    else
      local xi = -a[i]
      local yi = b[i]
      local gcd = getgcd(xi, yi)
      table.insert(x2, xi / gcd)
      table.insert(y2, yi / gcd)
    end
  end
end
-- local idx1 = {}
-- for i = 1, #x1 do
--   idx1[i] = i
-- end
-- local idx2 = {}
-- for i = 1, #x2 do
--   idx2[i] = i
-- end
-- table.sort(idx1, function(aa, bb)
--   if x1[aa] ~= x1[bb] then
--     return x1[aa] < x1[bb]
--   else
--     return y1[aa] < y1[bb]
--   end
-- end)
--
-- table.sort(idx2, function(aa, bb)
--   if y2[aa] ~= y2[bb] then
--     return y2[aa] < y2[bb]
--   else
--     return x2[aa] < x2[bb]
--   end
-- end)

local pow2 = {1}
for i = 1, 200010 do
  pow2[i + 1] = (pow2[i] * 2) % mod
end

local ret = badd(pow2[xzero + 1] - 1, pow2[yzero + 1] - 1)
ret = (ret + 1) % mod


local m1 = {}
for i1 = 1, #x1 do
  local xs, ys = tostring(x1[i1]), tostring(y1[i1])
  if not m1[xs] then m1[xs] = {} end
  if not m1[xs][ys] then
    m1[xs][ys] = 1
  else
    m1[xs][ys] = m1[xs][ys] + 1
  end
end

local m2 = {}
for i2 = 1, #x2 do
  local xs, ys = tostring(x2[i2]), tostring(y2[i2])
  if not m2[xs] then m2[xs] = {} end
  if not m2[xs][ys] then
    m2[xs][ys] = 1
  else
    m2[xs][ys] = m2[xs][ys] + 1
  end
end
for k1, v in pairs(m1) do
  for k2, c1 in pairs(v) do
    local c2 = nil
    if m2[k2] then c2 = m2[k2][k1] end
    if c2 then
      local val = pow2[c1 + c2 + 1]
      local rm = bmul(pow2[c1 + 1] - 1, pow2[c2 + 1] - 1)
      ret = bmul(ret, bsub(val, rm))
      m2[k2][k1] = nil
    else
      ret = bmul(ret, pow2[c1 + 1])
    end
  end
end
for k1, v in pairs(m2) do
  for k2, c in pairs(v) do
    ret = bmul(ret, pow2[c + 1])
  end
end

print(bsub(ret, 1))