There are n Imperial stormtroopers on the field. The battle field is a plane with Cartesian coordinate system. Each stormtrooper is associated with his coordinates (x, y) on this plane.

Han Solo has the newest duplex lazer gun to fight these stormtroopers. It is situated at the point (x

Your task is to determine what minimum number of shots Han Solo needs to defeat all the stormtroopers.

The gun is the newest invention, it shoots very quickly and even after a very large number of shots the stormtroopers don't have enough time to realize what's happening and change their location.

Input

The first line contains three integers n, x0 и y0 (1 ≤ n ≤ 1000,  - 104 ≤ x0, y0 ≤ 104) — the number of stormtroopers on the battle field and the coordinates of your gun.

Next n lines contain two integers each xi, yi ( - 104 ≤ xi, yi ≤ 104) — the coordinates of the stormtroopers on the battlefield. It is guaranteed that no stormtrooper stands at the same point with the gun. Multiple stormtroopers can stand at the same point.

Output

Print a single integer — the minimum number of shots Han Solo needs to destroy all the stormtroopers.

Examples

Input

4 0 0
1 1
2 2
2 0
-1 -1

Output

2

Input

2 1 2
1 1
1 0

Output

1

Note

Explanation to the first and second samples from the statement, respectively:

思路：

用pair和set結合，存斜率的分子和分母。最後輸出size。注意細節，座標軸要特殊考慮，要和四個象限分開，遇到座標軸的題目要注意，還有gcd輸入不能輸入兩個0。

ac程式碼：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<set>
#include<iostream>
#include<map>
#include<stack>
#include<cmath>
#include<algorithm>
#define ll long long
#define mod 10003
#define eps 1e-8
using namespace std;
pair<int,int> p1;
set<pair<int,int> > ss;
ll gcd(ll a,ll b)
{
	if(b==0)
	return a;
	else
	return gcd(b,a%b);
}
int main()
{
	int n,x,y;
	scanf("%d%d%d",&n,&x,&y);
	for(int i=0;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		int tx=x-u;
		int ty=y-v;
		if(tx==0&&ty==0)
		{
			p1.first=0;
			p1.second=0;
			ss.insert(p1);
			continue;
		}
		else if(tx==0)
		{
			p1.first=0;
			p1.second=1;
			ss.insert(p1);
			continue;
		}
		else if(ty==0)
		{
			p1.first=1;
			p1.second=0;
			ss.insert(p1);
			continue;
		}
		int g=gcd(abs(tx),abs(ty));
		tx/=g;
		ty/=g;
		if(tx<0&&ty<0)
		{
			p1.first=-tx;
			p1.second=-ty;
			ss.insert(p1);
		}
		
		else if(tx<0&&ty>0)
		{
			p1.first=tx;
			p1.second=ty;
			ss.insert(p1);
		}
		else if(ty<0&&tx>0)
		{
			p1.first=-tx;
			p1.second=-ty;
			ss.insert(p1);
		}
		else
		{
			p1.first=tx;
			p1.second=ty;
			ss.insert(p1);
		}
	}
	cout<<ss.size()<<endl;
	return 0;
}