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: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(auto i=0; i<n ; ++i)
#define REP(i,a,b) for(auto i=a; i<b; ++i)
#define all(v) (v).begin() , (v).end()
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vc = vector<char>;
using vs = vector<string>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
using vvc = vector<vector<char>>;
using pii = pair<int , int>;
using pll = pair<long long , long long>;

//プロトタイプ宣言
void faster(void);
template <class T>
T gcd(T , T);
template <class T>
T get_digent(T);
template <class T>
T POW(T , T);


int main(){
  faster();
  ll n;
  cin >> n;
  vl x(n);
  rep(i,n) cin >> x[i];
  sort(all(x));
  ll _min=1e9;
  REP(i,x[0],x[n-1]+1){
    ll sum = 0;
    rep(j,n){
      sum += abs(x[j]-i)*abs(x[j]-i);
    }
    _min = min(_min , sum);
  }
  cout << _min ;
}


























//同期解除
void faster(void){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}

//ユークリッドの互除法
template <class T>
T gcd(T a, T b) {
  return (b ? gcd(b, a%b) : a);
} 

//桁数取得
template <class T>
T get_digent(auto x){
  string s;
  s = to_string(x);
  ll digent = s.size();
  return digent;
} 

//べき乗計算
template <class T>
T POW(T x, T y){
  if(y==0) return 1;
  if(y%2 == 1){
    ll temp = POW(x,y-1);
    return (temp*x);
  }
  else{
    ll temp = POW(x,y/2);
    return temp*temp;
  }
}