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 Counter
import math

n = int(input())
ab = []
mod = 10**9+7

num_ab0 = 0
num_a0 = 0
num_b0 = 0

for i in range(n):
    A, B = map(int, input().split())
    if A == 0 and B == 0:
        num_ab0 += 1
    elif A == 0:
        num_a0 += 1
    elif B == 0:
        num_b0 += 1
    else:
        g = math.gcd(A, B)
        if A < 0:
            A = -A
            B = -B
        ab.append((A//g, B//g))

c = Counter(ab)

total = 2 ** num_a0 + 2 ** num_b0 - 1
for k, v in c.items():
    if k[1] < 0:
        num = c.get((-k[1], k[0]))
    else:
        num = c.get((k[1], -k[0]))

    if num:
        if k[1] > 0:
            total *= (2 ** v + 2 ** num - 1)
    else:
        total *= 2 ** v

print((total - 1 + num_ab0) % mod)
