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: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;


//System.out.println();
public class Main implements Runnable { //クラス名はMain

    public static void main(String[] args) {
    	Thread.setDefaultUncaughtExceptionHandler((t,e)->System.exit(1));
        new Thread(null, new Main(), "", 16 * 1024 * 1024).start(); //16MBスタックを確保して実行
    }

    public void run() {
    	try {
	    	FastScannerMain3 sc = new FastScannerMain3(System.in);
	        int N = sc.nextInt();
	        int[] X = new int[100];

	        for(int i =0 ; i< N; i ++) {
	        	X[i] = sc.nextInt();
	        }

	        int[] sum = new int[100];
	        for(int p =0 ; p< 100; p ++) {
	        	sum[p] = 0;
	        	for(int k =0 ; k < N; k ++) {
	        		sum[p] += Math.abs(X[k] - (p+1)) * Math.abs((X[k] - (p+1)));
	        		//System.out.println(p + " , " + k + " " + Math.abs(X[k] - (p+1)) * Math.abs((X[k] - (p+1))));
		        }
	        	//System.out.println(sum[p]);
	        }

	        //System.out.println(Arrays.toString(sum));
	        System.out.println(Arrays.stream(sum).min().getAsInt());


    	}catch(Exception e) {
    		e.printStackTrace();
    	}

    }
}


//高速なScanner
class FastScannerMain3 {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScannerMain3(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}