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 <stdio.h>

int MIN(int a, int b)
{ return a < b ? a : b ;}

int SQU(int a)
{ return a*a;}

int main()
{
  int N,i,j,sum,min;
  int x[101] = {0};

  scanf("%d",&N);

  for(i=0;i<N;i++)
    scanf("%d",&x[i]);

  for(i=0; i<N; i++) /* 座標x[0]で集会を開く時の疲労度の合計 */
    sum += SQU(x[i]-x[0]);
  min = sum;

  for(i=x[0]; i<= x[N-1]; i++) /*座標 i で集会を開くときの疲労度の合計 */
    {
    sum = 0;
      for(j=0; j<N; j++){
        sum += SQU(x[j]-i);
      }
    min = MIN(min,sum);
    }

  printf("%d", min);
  return 0;
}