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
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
//const ll MOD=998244353;

long long gcd(long long a, long long b){
  a=abs(a);
  b=abs(b);
  if(a<b) swap(a,b);
  long long r;
  while(b){
    r=a%b;
    a=b;
    b=r;
  }
  return a;
}

long long modpow(long long a, long long n){
  long long r = 1;
  while (n) {
    if(n%2) r=r*a%MOD;
    a=a*a%MOD;
    n/=2;
  }
  return r;
}


int main(){
  int N; cin>>N;
  int Z=0;
  map<pair<ll,ll>,pair<ll,ll>> mp;
  rep(i,N){
    ll A,B; cin>>A>>B;
    if(A==0&&B==0) Z++;
    else if(A==0){
      mp[{0,0}].first++;
    }else if(B==0){
      mp[{0,0}].second++;
    }else{
      ll g=gcd(A,B);
      A=A/g;
      B=B/g;
      if(B<0){
        A=-A;
        B=-B;
      }
      if(A<0) mp[{B,-A}].first++;
      else mp[{A,B}].second++;
    }
  }
  ll ans=1;
  for(auto m:mp){
    ll tmp=modpow(2,m.second.first)+modpow(2,m.second.second)-1;
    ans=(ans*tmp)%MOD;
  }
  ans=(ans-1+Z+MOD)%MOD;
  cout<<ans<<endl;
  return 0;
}