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>
#define FOR(i, k, n) for(int i = (k); i < (n); i++)
#define FORe(i, k, n) for(int i = (k); i <= (n); i++)
#define FORr(i, k, n) for(int i = (k)-1; i > (n); i--)
#define FORre(i, k, n) for(int i = (k)-1; i >= (n); i--)
#define REP(i, n) FOR(i, 0, n)
#define REPr(i, n) FORre(i, n, 0)
#define ALL(x) (x).begin(), (x).end()
#define ALLr(x) (x).rbegin(), (x).rend()
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
using namespace std;
using ll = long long;

const int INF = 1001001001;

int main(void){
  int n;
  cin >> n;
  vector<int> x(n);
  REP(i, n) cin >> x[i];
  int ans = INF;
  FORe(p, 1, 100){
    int sum = 0;
    REP(i, n) sum += (x[i]-p)*(x[i]-p);
    chmin(ans, sum);
  }
  cout << ans << endl;
  return 0;
}