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;
using ll = long long;
using Graph = vector<vector<int>>;
#define rep(i,m,n) for(int (i)=(m);(i)<(n);++(i))
#define rREP(i,m,n) for(int (i)=(n)-1;(i)>=(m);--(i))
#define all(x) (x).begin(),(x).end()
#define out(y,x,h,w) (y)<0||(x)<0||(y)>=(h)||(x)>=(w)
constexpr int INF = 1000000001;
constexpr ll mod = 1000000007;
constexpr double PI = 3.1415926535897932;
template<class T>inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T>inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
inline void init() {
	cin.tie(nullptr); cout.tie(nullptr);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(15);
}

int main() {
	init();

	int N; cin >> N;
	vector<int>X(N);
	rep(i, 0, N)cin >> X[i];
	int ans = INF;
	rep(i, 0, 101) {
		int cnt = 0;
		rep(j, 0, N) {
			int x = X[j] - i;
			cnt += x * x;
		}
		chmin(ans, cnt);
	}
	cout << ans << "\n";

	return 0;
}