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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::input;
use std::collections::HashMap;

// // a,b の最大公約数と、ax + by = gcd(a,b) となる x,y を求める
// fn extgcd(a: i64, b: i64) -> (i64, i64, i64) {
//     if b != 0 {
//         let (g, x, y) = extgcd(b, a % b);
//         (g, y, x - (a / b) * y)
//     } else {
//         (a, 1, 0)
//     }
// }

// // ax ≡ gcd(a, m) (mod m) となる x を求める
// // a, m が互いに素ならば、関数値は mod m での a の逆数となる
// fn mod_inverse(a: i64, m: i64) -> i64 {
//     let (_, x, _) = extgcd(a, m);
//     return (x % m + m) % m;
// }

// // 組み合わせ (r<10^7, n,r<m, m:素数)
// fn nCr(n: i64, mut r: i64, m: i64) -> i64 {
//     if n < r {
//         return 0;
//     }
//     if n - r < r {
//         r = n - r;
//     }
//     let mut ret = 1;
//     let mut i = 0;
//     while i < r {
//         ret = ret * (n - i) % m;
//         i += 1;
//     }
//     let mut imul = 1;
//     i = 0;
//     while i < r {
//         imul = imul * (i + 1) % m;
//         i += 1;
//     }
//     ret = ret * mod_inverse(imul, m) % m;
//     return ret;
// }

/*
 * 計算量は O(log(n)) で次式の計算結果を返す
 * x**n (mod mod)
 * つまり，xを乗して，modで割った余りを返す．
 * ex) mod_pow(2, 3, 3) = 2
 * 素数pに対して x**p = x (mod p) が成り立つ
*/
fn mod_pow(x: i64, n: i64, m: i64) -> i64 {
    if n == 0 {
        return 1;
    }
    let mut res = mod_pow(x * x % m, n / 2, m);
    if n & 1 == 1 {
        res = res * x % m;
    }
    return res;
}

#[proconio::fastout]
fn main() {
    input! {
        n: usize,
        ab: [(i64, i64); n],
    }
    let m = 1000000007;
    let mut map: HashMap<num_rational::Ratio<i64>, (i64, i64)> = HashMap::new();
    let mut zero = 0;
    let mut b_zero = 0;
    for (a, b) in ab {
        if b == 0 {
            if a == 0 {
                zero += 1;
            } else {
                b_zero += 1;
            }
            continue;
        }
        let c;
        let d;
        let is_positive;
        if a * b >= 0 {
            c = a.abs();
            d = b.abs();
            is_positive = true;
        } else {
            c = b.abs();
            d = a.abs();
            is_positive = false;
        }
        let f = num_rational::Ratio::new(c, d);
        map.insert(
            f,
            if !map.contains_key(&f) {
                if is_positive {
                    (1, 0)
                } else {
                    (0, 1)
                }
            } else {
                if is_positive {
                    (map[&f].0 + 1, map[&f].1)
                } else {
                    (map[&f].0, map[&f].1 + 1)
                }
            },
        );
    }
    let mut ans = 1;
    for key in map.keys() {
        // println!("{} {} {}", key, map[key].0, map[key].1);
        if key == &num_rational::Ratio::new(0, 1) {
            continue;
        }
        let mut v = 1;
        v += mod_pow(2, map[key].0, m) - 1;
        v += mod_pow(2, map[key].1, m) - 1;
        ans *= v;
    }
    if b_zero > 0 {
        let mut v = 1;
        let key = num_rational::Ratio::new(0, 1);
        if map.contains_key(&key) {
            v += mod_pow(2, map[&key].0, m) - 1;
        }
        v += mod_pow(2, b_zero, m) - 1;
        ans *= v;
    }
    ans -= 1;
    ans += zero;
    println!("{}", ans);
}
*/

fn main() {
    let exe = "/tmp/binCB65E195";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";