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 int long long

#define ii pair<int,int>
#define fi first
#define sc second
#define lwb lower_bound
#define upb upper_bound
#define all(x) (x).begin(),(x).end()

#define sqr(x) ((x)*(x))

signed main() {
#ifdef _DEBUG
  // freopen("in", "r", stdin);
#endif
  ios::sync_with_stdio(0); cin.tie(0);
  int n; cin >> n;
  vector<int> a(n);
  int avr = 0;
  for (auto &i : a) cin >> i, avr += i;
  avr /= n;
  int lans = 0, rans = 0;
  for (auto &i : a) {
    lans += sqr(abs(i - avr));
    rans += sqr(abs(i - avr - 1));
  }
  cout << min(lans, rans) << '\n';
}