Number of Parallelograms

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n points on a plane. All the points are distinct and no three of them lie on the same line. Find the number of parallelograms with the vertices at the given points.

Input

The first line of the input contains integer n (1 ≤ n ≤ 2000) — the number of points.

Each of the next n lines contains two integers (x_i, y_i) (0 ≤ x_i, y_i ≤ 10⁹) — the coordinates of the i-th point.

Output

Print the only integer c — the number of parallelograms with the vertices at the given points.

Example

input

4 0 1 1 0 1 1 2 0

output

1 题解：平行四边形的中点是确定的，那么根据两两边的中点，如果有重合那么可以组成平行四边形； 代码：

import java.util.Arrays;import java.util.Comparator;import java.util.Scanner;public class 平行四边形数 {    static Scanner cin = new Scanner(System.in);    public static void main(String[] argvs){        int N = cin.nextInt();        Point[] a = new Point[N];        for(int i = 0; i < N; i++){            a[i] = new Point();            a[i].x = cin.nextDouble();            a[i].y = cin.nextDouble();            //a[i].Put();        }        Point[] b = new Point[N * (N + 1)/2];        int tp = 0;        for(int i = 0; i < N*(N + 1)/2; i++)                b[i] = new Point();        for(int i = 0; i < N; i++){            for(int j = i + 1; j < N; j++){                if(a[i].x == a[j].x && a[i].y == a[j].y)                    continue;                b[tp++] = Point.getp(a[i], a[j]);            }        }                Arrays.sort(b, 0, tp, new PointComparator());//        System.out.println("tp = " + tp);        int ans = 0, cnt = 1;        for(int i = 1; i < tp; i++){        //    System.out.println(b[i].x + " " + b[i].y);            if(b[i].x == b[i - 1].x && b[i].y == b[i - 1].y){                cnt++;            }            else{                ans += cnt * (cnt - 1) / 2;                cnt = 1;            }        }        System.out.println(ans);    }}class Point{    static Scanner cin = new Scanner(System.in);    public double x, y;    public void Put(){        x = cin.nextInt();        y = cin.nextInt();    //    System.out.println("***" + this.x + " " + this.y);    }    static Point getp(Point a, Point b){        Point c = new Point();        c.x = (a.x + b.x) / 2;        c.y = (a.y + b.y) / 2;    //    System.out.println("***" + c.x + " " + c.y);        return c;    }}class PointComparator implements Comparator{    public int compare(Object x, Object y){        Point a = (Point)x;        Point b = (Point)y;            if(a.x != b.x){                if(a.x <= b.x)                    return -1;                else                    return 1;            }            else{                if(a.y <= b.y)                    return -1;                else                    return 1;            }    }}