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
#写経
#https://atcoder.jp/contests/abc168/submissions/13414936
from math import gcd
from collections import Counter
mod = 10**9 + 7
N,*AB = map(int,open(0).read().split())

def std(a,b):
    """
    a,bを既約にする
    """
    if a == 0:
        return(0,int(b!=0))
    g = gcd(a,b)
    a,b = a// g, b//g
    return(a,b) if a > 0 else (-a,-b)

C = Counter(std(a, b) for a, b in zip(*[iter(AB)] * 2))
def resolve():
    ans = 1
    cnt = 0
    for (a,b), v in C.items():
        if b > 0:
            if (b,-a) in C:
                ans *= -1 + pow(2,v,mod) + pow(2,C[(b,-a)], mod) #仲の悪い組み合わせが一緒に入らないように，別々に入れるか入れないかを判定 空集合は共通しているので引く
                ans %= mod
            else:
                cnt += v
        elif (-b,a) not in C:
            cnt += v
    ans *= pow(2,cnt,mod)
    ans += C[(0,0)] - 1
    print(ans%mod)
resolve()