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)

/*
confirm 0LL and 1LL
confirm cornercases such as 0
confirm times of cin < 10^6
*/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll, ll>;
using Pld = pair<ld, ld>;
using Vec = vector<ll>;
using VecP = vector<P>;
using VecB = vector<bool>;
using VecC = vector<char>;
using VecD = vector<ld>;
using VecS = vector<string>;
using Graph = vector<VecP>;
template <typename T>
using Vec1 = vector<T>;
template <typename T>
using Vec2 = vector<Vec1<T> >;
#define REP(i, m, n) for(ll (i) = (m); (i) < (n); ++(i))
#define REPN(i, m, n) for(ll (i) = (m); (i) <= (n); ++(i))
#define REPR(i, m, n) for(ll (i) = (m)-1; (i) >= (n); --(i))
#define REPNR(i, m, n) for(ll (i) = (m); (i) >= (n); --(i))
#define rep(i, n) REP(i, 0, n)
#define repn(i, n) REPN(i, 1, n)
#define repr(i, n) REPR(i, n, 0)
#define repnr(i, n) REPNR(i, n, 1)
#define all(s) (s).begin(), (s).end()
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
template <typename T>
bool chmax(T &a, const T b){if(a < b){a = b; return true;} return false;}
template <typename T>
bool chmin(T &a, const T b){if(a > b){a = b; return true;} return false;}
template <typename T>
ll pow2(const T n){return (1LL << n);}
template <typename T>
void cosp(const T n){cout << n << ' ';}
template <typename T>
void co(const T n){cout << n << '\n';}
template <typename T1, typename T2>
void co(pair<T1, T2> p){cout << p.fs << ' ' << p.sc << '\n';}
template <typename T>
void co(const Vec1<T> &v){for(T i : v) cosp(i); cout << '\n';}
template <typename T>
void co(initializer_list<T> v){for(T i : v) cosp(i); cout << '\n';}
template <typename T>
void ce(const T n){cerr << n << endl;}
void sonic(){ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);}
void setp(const ll n){cout << fixed << setprecision(n);}
const ll INF = 1e9+1;
const ll LINF = 1e18+1;
const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const ld PI = acos(-1);
const ld EPS = 1e-11;

int main(void){
	ll n;
	cin >> n;
	Vec x(n);
	rep(i, n) cin >> x[i];

	ll ans = LINF;
	rep(i, 105){
		ll sum = 0;
		rep(j, n) sum += (x[j]-i)*(x[j]-i);
		chmin(ans, sum);
	}
	co(ans);

	return 0;
}