Problem:
Score: 500 points

Problem Statement

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.

Sample Input 1

3
1 2
-1 1
2 -1

Sample Output 1

5

There are five ways to choose the set of sardines, as follows:

The 1-st

The 1-st and 2-nd

The 2-nd

The 2-nd and 3-rd

The 3-rd

Sample Input 2

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

Sample Output 2

479

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

#![allow(non_snake_case)]
use num::integer::gcd;
use proconio::input;
use std::collections::HashMap;
const M: i64 = 1000000007_i64;
fn main() {
    input! {
        n:usize,
      mut  AB:[(i64,i64);n]
    }
    // 鰯(Ai,Bi)と(Aj,Bj)について、Ai*Aj + Bi*Bj =0になる組み合わせは不可。
    // (0,0)の鰯の数を x とすると、これらは一匹のみの場合のみ追加可能。
    // 両辺を Aj * Bi で割ると、Ai/Bi + Bj/Aj = 0となる。
    // 逆数にして -1 をかけた鰯とは組み合わせられないが、他ならOK。
    // 鰯について、既約分数で表現すれば、Ai/Biは一意に定まる。
    let mut group1 = HashMap::new();
    // modPow(2,x)
    let mut mp2 = vec![1i64; (n as usize) + 1];
    for i in 0..(n as usize) {
        mp2[i + 1] = mp2[i] * 2 % M;
    }
    //let mut group2 = HashMap::new();
    let mut AB: Vec<(i64, i64)> = AB.into_iter().filter(|&(a, b)| a != 0 && b != 0).collect();
    let zero = (n - AB.len()) as i64;
    AB = AB
        .into_iter()
        .map(|(a, b)| {
            let g = gcd(a, b);
            (a / g, b / g)
        })
        .collect();
    for &x in AB.iter() {
        *group1.entry(x).or_insert(0) += 1;
    }
    let mut ans = 1;

    let mut get_pararell = |a: i64, b: i64| {
        let mut res = 0;
        if let Some(v) = group1.get(&(a, b)) {
            res += v;
            group1.remove(&(a, b));
        }
        if let Some(v) = group1.get(&(-a, -b)) {
            res += v;
            group1.remove(&(-a, -b));
        }
        res
    };
    //    println!("{:?}", group1);
    for &(a, b) in AB.iter() {
        let pararells = get_pararell(a, b);
        let crosses = get_pararell(b, -a);
        // println!("({},{}), p:{}, c:{}", a, b, pararells, crosses);
        ans = ans * ((mp2[pararells] + mp2[crosses] - 1) % M) % M;
    }
    ans = (ans + zero - 1) % M;
    println!("{}", ans);
}

// また、bの符号に一致する様に計算する。

fn yakubun(a: i64, b: i64) -> (i64, i64) {
    if b == 0 {
        return (-1, 0);
    }
    if a == 0 {
        return (0, 1);
    }
    let x = if b > 0 { gcd(a, b) } else { -gcd(a, b) };
    return (a / x, b / x);
}
//
//fn modPow(x: i64, a: i64, mp2: &Vec<i64>) -> i64 {
//    if x == 2 && (a as usize) < mp2.len() {
//        return mp2[a as usize];
//    }
//    if a == 1 {
//        x
//    } else {
//        if a % 2 != 0 {
//            (modPow(x, a - 1, &mp2) * x) % M
//        } else {
//            let t = modPow(x, a / 2, &mp2);
//            t * t % M
//        }
//    }
//}
////fn modInv(x: i64) -> i64 {
////    modPow(x, M - 2)
//}