题干：

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.

Examples

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

题目大意：

    定义一个y()函数，然后定义一个s(x)函数，问我们s(x)函数上有多少个不是180°的倾角。

解题报告：

      我们假象y(x)不是那样定义的，它就是简单的一次函数，那么n个一次函数相加肯定是一次函数，都是180°的倾角（直的），那么问题就出现在y(x)<0时，y(x)=0,也就是它本身应该加上一个小于0的数结果加上0了，所以就会有弯曲。也就是记录与x轴有多少不同的交点。

      这题要是直接用set存，然后需要开long double 记录斜率，并且选择合适的编译器，才可以ac。。。不然就卡精度

     更好的方法是直接以最简分数的方式存啊。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<set>
#define ll long long
#define mp make_pair
using namespace std;
struct Node {
	ll b,k;
} node[100000 + 5]; 
set<pair<ll,ll> > st;
int main ()
{
	ll n,k,b;
	cin>>n;
	for(int i = 1; i<=n; i++) {
		scanf("%lld%lld",&k,&b);
		if(k == 0) continue;
		if(k<0&&b<0) k=-k,b=-b;
		if(b == 0) {
			st.insert(mp(0,0));
			continue;
		}
		if(k<0&&b>0) k=-k,b=-b;
		st.insert(mp(k/__gcd(abs(k),abs(b)),b/__gcd(abs(k),abs(b)))); 
	}
	cout << st.size() << endl;
	return 0 ;
}