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)

#define _USE_MATH_DEFINES
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <clocale>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <queue>
#include <regex>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i, s, n) for(int i = (s), i##_len=(n); i < i##_len; ++i)
#define FORS(i, s, n) for(int i = (s), i##_len=(n); i <= i##_len; ++i)
#define VFOR(i, s, n) for(int i = (s); i < (n); ++i)
#define VFORS(i, s, n) for(int i = (s); i <= (n); ++i)
#define REP(i, n) FOR(i, 0, n)
#define REPS(i, n) FORS(i, 0, n)
#define VREP(i, n) VFOR(i, 0, n)
#define VREPS(i, n) VFORS(i, 0, n)
#define RFOR(i, s, n) for(int i = (s), i##_len=(n); i >= i##_len; --i)
#define RFORS(i, s, n) for(int i = (s), i##_len=(n); i > i##_len; --i)
#define RREP(i, n) RFOR(i, n, 0)
#define RREPS(i, n) RFORS(i, n, 0)
#define ALL(v) (v).begin(), (v).end()
#define SORT(v) sort(ALL(v))
#define RSORT(v) sort(ALL(v), greater<decltype(v[0])>())
#define SZ(x) ((int)(x).size())
#define REV(x) reverse(ALL(x))
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define MT make_tuple
#define BIT(n) (1LL<<(n))
#define UNIQUE(v) v.erase(unique(ALL(v)), v.end())

using ld = long double;
using ll = long long;
using ui = unsigned int;
using ull = unsigned long long;
using Pi_i = pair<int, int>;
using Pll_ll = pair<ll, ll>;
using VB = vector<bool>;
using VC = vector<char>;
using VD = vector<double>;
using VI = vector<int>;
using VLL = vector<ll>;
using VS = vector<string>;
using VSH = vector<short>;
using VULL = vector<ull>;

const int MOD = 1000000007;
const int INF = 1000000000;
const int NIL = -1;
const double EPS = 1E-10;

template<class T, class S>
bool chmax(T &a, const S &b){
    if(a < b){
        a = b; return true;
    }
    return false;
}
template<class T, class S>
bool chmin(T &a, const S &b){
    if(b < a){
        a = b; return true;
    }
    return false;
}


int main(){
    int N; cin >> N;
    VI X(N);
    REP(i, N) cin >> X[i];
    int ans = INF;
    FORS(i, 1, 100){
        int cost = 0;
        REP(j, N) cost += (X[j]-i)*(X[j]-i);
        chmin(ans, cost);
    }
    cout << ans << endl;
    return 0;
}