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++ (GCC 9.2.1) (judge verdict: Accepted)

/*
    Author: Aryan Yadav
*/
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define mod 1000000007
#define f(i, a, b) for (long long i = a; i < b; i++)
#define rep(i, a, b) for (long long i = a; i <= b; i++)
#define fm(i, a, b) for (long long i = a; i > b; i--)
#define repm(i, a, b) for (long long i = a; i >= b; i--)
#define vi vector<int>
#define vl vector<long long>
#define vs vector<string>
#define pqmaxi priority_queue<int>
#define pqmaxl priority_queue<ll>
#define pqmini priority_queue<int, vi, greater<int>>
#define pqminl priority_queue<ll, vl, greater<ll>>
#define all(x) x.begin(), x.end()
#define test     \
    ll test;     \
    cin >> test; \
    while (test--)
#define pb(i) push_back(i)
#define mp(i, j) make_pair(i, j)
#define fast                     \
    ios::sync_with_stdio(false); \
    cin.tie(NULL)

int main()
{
    fast;
    ll n;
    cin >> n;
    vl v(n);
    f(i, 0, n)
    {
        cin >> v[i];
    }
    ll min1 = mod;
    ll max1 = -mod;
    f(i, 0, n)
    {
        min1 = min(min1, v[i]);
        max1 = max(max1, v[i]);
    }
    ll ans = mod;
    f(j, min1, max1 + 1)
    {
        ll cst = 0;
        f(i, 0, n)
        {
            cst += (v[i] - j) * (v[i] - j);
        }
        ans = min(ans, cst);
    }
    cout << ans;
}