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.

Examples

Input

3
1 2
-1 1
2 -1


Output

5


Input

10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4


Output

479
from collections import defaultdict
from math import gcd as g
n = int(input())
mod = 10**9+7
sardines = set()
Ddic = defaultdict(int)
ans = 1
zcount = 0
for i in range(n):
  a,b = map(int,input().split())
  if a == 0 and b == 0:
    zcount += 1
    continue

  if a == 0:
    Ddic[(0,1)] += 1
    sardines.add((0,1))
    continue

  if b == 0:
    sardines.add((1,0))
    Ddic[(1,0)] += 1
    continue

  A,B = abs(a),abs(b)
  G = g(A,B)
  a //= G
  b //= G

  if b < 0:
    a = -a
    b = -b

  Ddic[(a,b)] += 1
  sardines.add((a,b))

for i in sardines:
  if i == (0,0) or (i[0] <= 0 and Ddic[(i[1],-i[0])] != 0):continue
  if i[0] <= 0:
    ans *= (2**Ddic[i] + 2**(Ddic[(i[1],-i[0])]))-1
  else:
    ans *= (2**Ddic[i] + 2**(Ddic[(-i[1],i[0])]))-1

  ans %= mod
print((ans+zcount-1)%mod)
