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: Wrong Answer)

#include<iostream> 
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <vector>

using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)



int main() {
	int i = 0,n, h = 0, p, ans = 0;
	double d ;
	cin >> n;
	vector<int>x(n);
	rep(i, n) {
		cin >> x.at(i);
		h += x.at(i);
	}
	p = h / n;
	if (h % n > 4) {
		p++;
	}
	rep(i, n) {
		d = (double)p - (double)x.at(i);
		ans += (int)d * (int)d;
	}
	cout << ans << endl;

	return 0;
}