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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	static void quick_sort(int[] d, int left, int right) {
		if (left >= right) {
			return;
		}
		int p = d[(left + right) / 2];
		int l = left, r = right;
		int tmp;
		while (l <= r) {
			while (d[l] < p) {
				l++;
			}
			while (d[r] > p) {
				r--;
			}
			if (l <= r) {
				tmp = d[l];
				d[l] = d[r];
				d[r] = tmp;
				l++;
				r--;
			}
		}
		quick_sort(d, left, r); // ピボットより左側をクイックソート
		quick_sort(d, l, right); // ピボットより右側をクイックソート
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int address[] = new int[N];
		for (int i = 0; i < N; i++) {
			address[i] = sc.nextInt();
		}
		if (N == 1) {
			System.out.println(address[0]);
		} else {

			quick_sort(address, 0, N - 1);
			int sum = 0;
			int min = 0;
			for (int i = 0; i < N; i++) {
				min += (address[i] - address[0]) * (address[i] - address[0]);
			}

			for (int i = address[0] + 1; i <= address[N - 1]; i++) {
				for (int j = 0; j < N; j++) {
					sum += (address[j] - i) * (address[j] - i);
				}
				if (sum < min) {
					min = sum;
				}
				sum = 0;
			}

			System.out.println(min);
		}
	}
}