CodeForces 514B Han Solo and Lazer Gun

最新推荐文章于 2018-11-24 20:55:21 发布

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此编程问题探讨了汉索罗使用激光枪消灭战场上的帝国风暴兵所需的最少射击次数。战场采用笛卡尔坐标系，汉索罗位于特定点上，可以沿着通过该点的任意直线射击。任务是确定消灭所有风暴兵所需的最小射击次数。

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.

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 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.

Output

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

Examples

Input

Output

Input

2 1 2
1 1
1 0

Output

Note

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

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

double a[1010];

int main(){
    int n,i,x,y;
    int xx,yy;
    int cnt,res;
    int flag;
    while(~scanf("%d %d %d",&n,&x,&y)){
    	cnt=res=0;
     	flag=0;
        for(i=0;i<n;i++){
            scanf("%d %d",&xx,&yy);
            double jx=xx-x;
            double jy=yy-y;
            if(!flag&&jx==0){
                flag=1;
                res++;
            }
            else if(jx!=0){
                a[cnt++]=jy/jx;
            }
        }
        sort(a,a+cnt);
        for(i=0;i<cnt;i++){
        	while(a[i]==a[i+1]&&i<cnt){
	        	i++;
	        }	
        	res++;
        }
        printf("%d\n",res);
    }
    return 0;
}