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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;
import java.util.HashMap;
import java.util.Objects;
import java.util.Map;
import java.util.Map.Entry;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        E solver = new E();
        solver.solve(1, in, out);
        out.close();
    }

    static class E {
        public void solve(int testNumber, Scanner in, PrintWriter out) {

            int num = in.nextInt();

            int bad_count = 0;

            Map<Pair, Long> map = new HashMap<>();

            for (int i = 0; i < num; i++) {
                long a = in.nextLong();
                long b = in.nextLong();

                if (a == 0 && b == 0) {
                    bad_count++;
                } else if (b == 0) {
                    Pair pair = new Pair(1l, 0l);
                    map.put(pair, map.getOrDefault(pair, 0L) + 1);
                } else if (a == 0) {
                    Pair pair = new Pair(0l, -1l);
                    map.put(pair, map.getOrDefault(pair, 0L) + 1);

                } else {
                    long g = gcd(a, b);
                    a = a / g;
                    b = b / g;
                    if (a < 0) {
                        a *= -1;
                        b *= -1;
                    }
                    Pair pair = new Pair(a, b);
                    map.put(pair, map.getOrDefault(pair, 0L) + 1);
                }
            }

            long ans = 1;

            while (!map.keySet().isEmpty()) {
                Pair key = map.entrySet().iterator().next().getKey();

                Pair pair;
                if (key.second < 0) {
                    pair = new Pair(-1 * key.second, key.first);
                } else {
                    pair = new Pair(key.second, -1 * key.first);
                }

                ans *= (Math.pow(2, map.get(key)) - 1) + (Math.pow(2, map.getOrDefault(pair, 0L)) - 1) + 1;
                ans %= 1000000007;

                if (map.containsKey(pair)) {
                    map.remove(pair);
                }
                map.remove(key);
            }
            ans = (ans + bad_count - 1) % 1000000007;

            out.append(String.valueOf(ans));
        }

        long gcd(long a, long b) {
            return b == 0 ? a : gcd(b, a % b);
        }

        class Pair implements Comparable<E.Pair> {
            public final Long first;
            public final Long second;

            Pair(Long first, Long second) {
                this.first = first;
                this.second = second;
            }

            public boolean equals(Object obj) {
                if (!(obj instanceof Pair))
                    return false;
                Pair pair = (Pair) obj;
                return pair.first.equals(first) && pair.second.equals(second);
            }

            public int hashCode() {
                return Objects.hash(first, second);
            }

            public int compareTo(Pair o) {
                int res = Long.compare(first, o.first);
                return res != 0 ? res : Long.compare(second, o.second);
            }

        }

    }
}