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 deque
from math import gcd
import sys

def input():
    return sys.stdin.readline().rstrip()

MOD = 10**9 + 7
n = int(input())
cnt = {}
zero_cnt = 0
ans = 1

for _ in range(n):
    a, b = map(int, input().split())
    if (a, b) == (0, 0):
        zero_cnt += 1
        continue
    g = gcd(abs(a), abs(b))
    if a == 0 or b == 0: g = max(abs(a), abs(b))
    a //= g
    b //= g

    rotate = 0
    while not (a > 0 and b >= 0):
        a, b = -b, a
        rotate += 1
    if (a, b) not in cnt:
        cnt[(a, b)] = [0, 0]

    cnt[(a, b)][rotate%2] += 1

for key in cnt:
    A, B = cnt[key]
    ans *= 1 + pow(2, A, MOD) - 1 + pow(2, B, MOD) - 1
    ans %= MOD

ans += zero_cnt
ans -= 1 #for empty set
print(ans % MOD)
