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;

#if DEBUG && !ONLINE_JUDGE
  #include "header.h"
#else
  #define debug(args...)
#endif

typedef vector<int> vi;
typedef pair<int,int> ii;
typedef long long lli;
typedef long double ld;

#define pb push_back
#define all(x) x.begin(),x.end()
#define SZ(x) (int)(x).size()
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define rep(i,a,b) for(auto i=(a);i<b;i++)

#define INF (int)1e9
#define EPS 1e-9
#define MOD 1000000007
void preprocess(void) {
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  preprocess();
  int n; cin>>n;
  vi x(n);
  rep(i,0,n) cin>>x[i];
  lli ans = 0;
  rep(i,0,n) ans += x[i];
  ans /= n;
  lli fans = MOD * 1LL * MOD;
  rep(k,ans-2,ans+2) {
    lli sum = 0;
    rep(i,0,n) sum += (x[i]-k)*1LL*(x[i]-k);
    fans = min(fans, sum);
  }
  cout<<fans<<endl;
}