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(int i=0, i##_len=(n); i<i##_len; ++i)
#define all(x) (x).begin(),(x).end()
using ll = long long;
string char_to_string(char val) {
  return string(1, val);
}
int char_to_int(char val) {
  return val - '0';
}
template<class T> inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template<class T> inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
int vector_finder(std::vector<ll> vec, int number) {
  auto itr = std::find(vec.begin(), vec.end(), number);
  size_t index = std::distance( vec.begin(), itr );
  if (index != vec.size()) { // 発見できたとき
    return 1;
  }
  else { // 発見できなかったとき
    return 0;
  }
}

int main() {
    int N; cin >> N;
    vector<int> X(N); REP(i, N) cin >> X[i];
    int res = pow(10, 8);
    int max_n = *max_element(all(X));
    REP(i, max_n) {
        int tmp = 0;
        REP(k, N) {
            tmp += pow((X[k] - (i + 1)), 2);
        }
        if(tmp < res) res = tmp;
    }
    cout << res << endl;
}