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(x₀,?y₀). In one shot it can can destroy all the stormtroopers, situated on some line that crosses point(x₀,?y₀).

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.

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 x_i, y_i (?-?10⁴?≤?x_i,?y_i?≤?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.

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

4 0 01 12 22 0-1 -1Output
2Input
2 1 21 11 0Output
1Note
Explanation to the first and second samples from the statement, respectively: 
題目大意：
一共有N個(gè)目標(biāo)，給出N個(gè)目標(biāo)點(diǎn)的坐標(biāo)，并且給出機(jī)槍的目標(biāo)。
這個(gè)機(jī)槍非常牛X，其開一槍，前后的人都會(huì)死。
問最少開多少槍，能夠?qū)⑺心繕?biāo)都打到。
思路：
問題尋找共性。
對(duì)于一條路徑上的所有目標(biāo)都有一個(gè)唯一的共性，那就是斜率。
那么我們對(duì)問題的斜率進(jìn)行處理即可。任務(wù)目標(biāo)就是統(tǒng)計(jì)斜率的個(gè)數(shù)。
K=（y-y0）/（x-x0）；
注意這個(gè)K需要用double類型。
過程用map維護(hù)一下第一次出現(xiàn)即可。
Ac代碼：
#include<stdio.h>#include<string.h>#include<map>using namespace std;int main(){    int n;    int x,y;    while(~scanf("%d%d%d",&n,&x,&y))    {        int flag1=0;        int flag2=0;        int output=0;        map<double ,int >s;        for(int i=0;i<n;i++)        {            int xx,yy;            scanf("%d%d",&xx,&yy);            if(xx==x)            {                flag1=1;                continue;            }            if(yy==y)            {                flag2=1;                continue;            }            double  k=(yy-y)*1.0/(xx-x)*1.0;            if(s[k]==0)            {                s[k]=1;                output++;            }        }        printf("%d/n",output+flag1+flag2);    }}