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 "algorithm"
#include "cmath"
#include "iomanip"
#include "iostream"
#include "string"
#include "vector"
#define rep(i, to) for (ll i = 0; i < (to); ++i)
#define rep1(i, to) for (ll i = 1; i <= (to); ++i)
#define all(vec) vec.begin(), vec.end()
#define unless(cond) if (!(cond))
using namespace std;
typedef long long ll;
template <typename T>
using V = vector<T>;

template <typename T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

template <typename T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}

template <typename T>
void drop(const T& x) {
  cout << x << endl;
  exit(0);
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  ll N;
  V<ll> xs(N);
  rep(i, N) cin >> xs[i];

  ll ret = 1e18;

  rep1(i, 100) {
    ll sum = 0;

    rep(i, N) { sum += pow(xs[i] - i, 2); }

    chmin(ret, sum);
  }

  cout << ret << endl;

  return 0;
}