問題描述:

As Valeric and Valerko were watching one of the last Euro Championship games in a sports bar, they broke a mug. Of course, the guys paid for it but the barman said that he will let them watch football in his bar only if they help his son complete a programming task. The task goes like that.

Let's consider a set of functions of the following form:

Let's define a sum of n functions y1(x), ..., yn(x) of the given type as functions(x) = y1(x) + ... + yn(x) for any x. It's easy to show that in this case the graph s(x)is a polyline. You are given n functions of the given type, your task is to find the number of angles that do not equal 180 degrees, in the graph s(x), that is the sum of the given functions.

Valeric and Valerko really want to watch the next Euro Championship game, so they asked you to help them.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of functions. Each of the following n lines contains two space-separated integer numbers ki, bi

 ( - 109 ≤ ki, bi ≤ 109) that determine the i-th function.

Output

Print a single number — the number of angles that do not equal 180 degrees in the graph of the polyline that equals the sum of the given functions.

Input

1
1 0

Output

1

Input

3
1 0
0 2
-1 1

Output

2

Input

3
-2 -4
1 7
-5 1

Output

3

題目題意:問我們s(x)函式上有多少個不是180°的傾角。

題目分析:我們假象y(x)不是那樣定義的，它就是簡單的一次函式，那麼n個一次函式相加肯定是一次函式，都是180°的傾角（直的），那麼問題就出現在y(x)<0時，y(x)=0,也就是它本身應該加上一個小於0的數結果加上0了，所以就會有彎曲。

也就是記錄與x軸有多少不同的交點

程式碼如下:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#define ll long long
using namespace std;

set<long double> M;
int main()
{
    M.clear();
    int n;
    scanf("%d",&n);
    for (int i=1;i<=n;i++) {
        ll k,b;
        scanf("%lld%lld",&k,&b);
        if (k==0) continue;
        else {
            long double ans=-(long double)b/k;
            if (M.count(ans)==0) M.insert(ans);
        }
    }
    printf("%d\n",M.size());
    return 0;
}