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;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
/*
*/

ll gcd(ll a,ll b){
  if(b == 0) return a;
  return gcd(b, a % b);
}
const ll p =  1000000007;
ll factorial(ll n){
   ll ans = 1;
   for(int i = n;i>=1;i--){
     ans *= i;
     ans %= p;
   }
   return ans;
}
using Graph = vector<vector<int>>;

ll kzyou(ll a,ll k){//aのk乗を返す
    ll zyou = 1;
    for(ll i = 1;i<=k;i++){
        zyou *= a;
    }
    return zyou;
}

int main(){
  int N;
  cin >> N;
  vector<int> X(N);
  rep(i,N){
    cin >> X[i];
  }
  sort(X.begin(),X.end());
  
  ll ans = p;

  for(int P = X[0];P <= X[N-1];P++){
    ll sum = 0;
    for(int i = 0;i<N;i++){
      sum += (X[i] - P) * (X[i] - P);
    }
    ans = min(ans,sum);
  }
  cout << ans << endl;
}