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>
#include<unordered_map>
using namespace std;
#define ll long long
#define all(x) (x).begin(),(x).end()
#define rep(i, n) for(ll i = 0; i < n; i++)
const ll INF = 100100100100;


ll combination(ll n, ll r){
    if(n < r) return 0;
    if(r == 0||r == n) return 1;
    else if(r == 1) return n;
    return (combination(n - 1, r - 1) + combination(n - 1, r));
}

bool checkKaibun(string str){
    string checkStr = str;
    reverse(checkStr.begin(), checkStr.end());
    if(str == checkStr) return true;
    else return false;
}

int keta(ll num){
    int digits = 0;
    while(num > 0){
        num /= 10;
        digits++;
    }
    return digits;
}


int main(){
    int N;
    cin >> N;
    vector<int> a(N);
    rep(i, N)cin >> a[i];
    ll ans = INF;
    for(int i = 1; i <= 99 ; i++){
        ll tmp = 0;
        for(int j = 0; j < N; j++){
            tmp += (a[j] - i) * (a[j] - i);
        }
        if(ans > tmp)ans = tmp;
    }
    cout << ans << endl;
    
}