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;

#define ll long long 
#define inf 0x3f3f3f3f
#define infll 0x3f3f3f3f3f3f3f3f
#define all(x) (x).begin (),(x).end()
#define f(i, a, b) for (int i=a; i<(signed)(b); ++i)
#define pc __builtin_popcount // counts number of 1's in bin(num)

const double pi = 2*acos(0.0);
//__gcd(n1, n2);
//cout << fixed;
//cout << setprecision(6); //fixes # of dp



int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

    int n;
    cin >> n;
    int x[n]; 
    int mid = 0;
    f(i, 0, n){
        cin >> x[i];
        mid += x[i];
    }
    mid /= n;
    mid += 1;
    int ans1 = 0;
    f(i, 0, n){
        ans1 += (mid - x[i])*(mid-x[i]);
    }

    mid -= 1;
    int ans2 = 0;
    f(i, 0, n){
        ans2 += (mid - x[i])*(mid-x[i]);
    }
    cout << min(ans1,ans2);
}