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
import math
def main():
    N = int(input())
    d = {}
    za = zb = zab = r = 0
    mod = 10**9 + 7
    for i in range(N):
        a, b = map(int, input().split())
        if a== 0 and b == 0:
            zab += 1
        elif b == 0:
            zb += 1
        elif a == 0:
            za += 1
        else:
            if a < 0:
                a, b = -a, -b
            x = math.gcd(abs(a), abs(b))
            d[(a//x, b//x)] = d.get((a//x, b//x), 0) + 1
    used = set()
    l = []
    for x in d:
        if x in used:
            continue
        a, b = x[0], x[1]
        used.add(x)
        if a * b > 0:
            t = (abs(b), -abs(a))
        else:
            t = (abs(b), abs(a))
        used.add(t)
        l.append((d[x], d.get(t, 0)))
    r = pow(2, za) + pow(2, zb) - 1
    for i in l:
        r *= (pow(2, i[0]) + pow(2, i[1]) - 1)
        r %= mod
    return (r - 1 + zab) % mod
print(main())
