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 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
    int i,a,b=0,c,A[101],d=0,e=0,f=0,g=0;
    scanf("%d",&i);
    for(a=0;a<i;a++)
    {
       scanf("%d",&A[a]);
    }
    for(a=0;a<(i-1);a++)
    {
        if(A[i]<=A[i+1])
            c=A[i+1];
        else
            c=A[i];
    }
    for(a=0;a<(i-1);a++)
    {
        if(A[i]<=A[i+1])
            d=A[i];
        else
            d=A[i+1];
    }
    for(a=d;a<=c;a++)
    {
        for(b=0;b<i;b++)
        {
            e=e+(A[i]-d)*(A[i]-d);
        }
        for(b=0;b<i;b++)
        {
            f=f+(A[i+1]-d)*(A[i+1]-d);
        }
    if(e<=f)
    {
        g=e;
    }
    else
    {
        g=f;
    }
    }
printf("%d",g);
return 0;
}