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.*;
import java.math.*;
import java.util.*;

class Pair{
    long n1, n2;
    public Pair(long n1, long n2)
    {
        this.n1 = n1;
        this.n2 = n2;
    }

    @Override
    public String toString() {
        return "Pair{" +
                "n1=" + n1 +
                ", n2=" + n2 +
                '}';
    }
}
public class Main{

    static long MOD = 1000000007L;
    static long [] fac;
    static int[][] dir = new int[][]{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    static long lMax = 0x3f3f3f3f3f3f3f3fL;
    static int iMax = 0x3f3f3f3f;
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        // Start writing your solution here. -------------------------------------
        //int t = sc.nextInt();
        int t = 1;
        while(t-- > 0)
        {
           int n = sc.nextInt();
           long a[] = new long[n], b[] = new long[n];
           long ans = 1L;
           long zzCnt = 0L, azCnt = 0L, bzCnt = 0L;
           HashMap <String, Pair> cnt = new HashMap();
           ArrayList <Pair> list = new ArrayList();
           for(int i = 0; i < n; i++)
           {
               a[i] = sc.nextLong();
               b[i] = sc.nextLong();
               if(a[i] == 0 && b[i] == 0) zzCnt++;
               else if(a[i] == 0) azCnt++;
               else if(b[i] == 0) bzCnt++;
               else
               {
                   long g = gcd(a[i], b[i]);
                   a[i] /= g;
                   b[i] /= g;
                   if((a[i] > 0 && b[i] > 0) || a[i] < 0 && b[i] < 0)
                   {
                       String code = Math.abs(a[i]) + "#" + Math.abs(b[i]);
                       if(cnt.get(code) == null)
                           cnt.put(code, new Pair(1, 0));
                       else
                           cnt.get(code).n1++;
                   }
                   else
                   {
                       String code = Math.abs(b[i]) + "#" + Math.abs(a[i]);
                       if(cnt.get(code) == null)
                           cnt.put(code, new Pair(0, 1));
                       else
                           cnt.get(code).n2++;
                   }
               }
           }
           for(Pair p : cnt.values()) list.add(p);
           list.add(new Pair(azCnt, bzCnt));
           for(Pair p : list)
               ans = ans * (quickPOW(2L, p.n1) + quickPOW(2L, p.n2) - 1) % MOD;
           ans = (ans + zzCnt - 1) % MOD;
           out.println(ans);
        }
        out.close();
    }
    public static long C(int n, int m)
    {
        if(m == 0 || m == n) return 1l;
        if(m > n || m < 0) return 0l;
        long res = fac[n] * quickPOW((fac[m] * fac[n - m]) % MOD, MOD - 2) % MOD;

        return res;
    }
    public static long quickPOW(long n, long m)
    {
        long ans = 1l;
        while(m > 0)
        {
            if(m % 2 == 1)
                ans = (ans * n) % MOD;
            n = (n * n) % MOD;
            m >>= 1;
        }
        return ans;
    }
    public static int gcd(int a, int b)
    {
        if(a % b == 0) return b;
        return gcd(b, a % b);
    }
    public static long gcd(long a, long b)
    {
        if(a % b == 0) return b;
        return gcd(b, a % b);
    }
    public static long solve(long x, long[] res){
        int n = res.length;
        long a = x / n;
        int b = (int)(x % n);
        return res[n - 1] * a + res[b];
    }
    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    //--------------------------------------------------------
}