CodeForces 514B

B. Han Solo and Lazer Gun

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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:

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

 ///咋看吓一跳，感觉暴力不科学啊，怎么可能知道一条直线上有
 ///多少点，所以就去计算每个点同炮的斜率，斜率一样，就在同
 ///同一条直线上。
 const int MAX_SIZE = 1E4 + ;

 int main()
 {
     //freopen("in.txt", "r", stdin);
     int n, i;
     int x, y, px, py;
     double kx, ky;
     bool tag;
     double k[MAX_SIZE];

     while(scanf("%d %d %d", &n, &x, &y) != EOF)
     {
         tag = false;
         int j = ;
         for(i = ; i < n; i++)
         {
             scanf("%d %d", &px, &py);
             kx = px - x;
             ky = py - y;
             if(kx == )
                 tag = true;
             else
                 k[j++] = ky / kx;
         }

         sort(k, k+j);
         int ans = (tag == true) ?  : ;

         for(i = ; i < j-; i++)
         {
             if(k[i] != k[i+])
             {
                 ans++;
             }
         }
         ans += (i == j-) ?  : ;
         cout << ans << endl;
     }
     return ;
 }

 //题目需要注意的地方是，在中间计算过程，要用double去存储横纵坐标的差值。