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: Accepted)

#include<stdio.h>
#include<math.h>
#include<string.h>
int main()
{
    int a,c=0,d=0,e,max,min,ji=0;
    scanf("%d",&a);
    int b[a];
    for(c=0;c<a;c++)
    {scanf("%d",&b[c]);
    d=d+(b[c]-a)*(b[c]-a);}
    max=b[0]; min=b[0];
    for(c=1;c<a;c++)
    {
        max=max>b[c]? max:b[c];
        min=min<b[c]? min:b[c];
    }
    for(;min<=max;min++)
    {
        ji=0;
        for(e=0;e<a;e++)
            ji=ji+(b[e]-min)*(b[e]-min);
        d=d<ji? d:ji;
    }
    printf("%d",d);
    return 0;
}