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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ABC156
{
	class C
	{
		static void Main(string[] args)
		{
			var N = int.Parse(Console.ReadLine());
			var Xn = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();

			var midpoint = Xn.Average();
			var upper = GetTotalDistance(Xn, Math.Floor(midpoint));
			var lower = GetTotalDistance(Xn, Math.Ceiling(midpoint));

			var result = GetEfficientPoint(upper, lower);
			Console.WriteLine(result);
		}

		private static double GetEfficientPoint(double upper, double lower)
		{
			if(upper > lower)
			{
				return lower;
			} else
			{
				return upper;
			}
		}

		private static double GetTotalDistance(int[] xn, double average)
		{
			double sum = 0;
			foreach (var x in xn)
			{
				sum += Math.Pow(x - average, 2);
			}

			return sum;
		}
	}
}