Problem:
Score : 300 points

Problem Statement

There are N people living on a number line.

The i-th person lives at coordinate X_i.

You are going to hold a meeting that all N people have to attend.

The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.

Find the minimum total points of stamina the N people have to spend.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq X_i \leq 100

Input

Input is given from Standard Input in the following format:

N
X_1 X_2 ... X_N

Output

Print the minimum total stamina the N people have to spend.

Sample Input 1

2
1 4

Sample Output 1

5

Assume the meeting is held at coordinate 2. In this case, the first person will spend (1 - 2)^2 points of stamina, and the second person will spend (4 - 2)^2 = 4 points of stamina, for a total of 5 points of stamina. This is the minimum total stamina that the 2 people have to spend.

Note that you can hold the meeting only at an integer coordinate.

Sample Input 2

7
14 14 2 13 56 2 37

Sample Output 2

2354

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ifor(i,n) for(i=0;i<n;i++)
#define ll long long
///vector<int>vec;
///sort(s.begin(),s.endl());
///sort(ara,ara+n);
int main()
{
     ios_base::sync_with_stdio(false);
     cin.tie(NULL);
    int i,n,t,p=INT_MAX,j=0,x,a=0,b,l=0,r=INT_MAX;
     string s,m;
     cin>>n;
     int ara[n];
    ifor(i,n){
        cin>>ara[i];
        j+=ara[i];
        l=max(l,ara[i]);
        r=min(r,ara[i]);
    }
    p=j/n;

    if(j%n!=0){
        if(l-p>r-p){
            p++;
        }
    }

    ifor(i,n){
        a+=(abs(p-ara[i]))*(abs(p-ara[i]));
    }
     cout<<a<<endl;

     return EXIT_SUCCESS;
}