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++14 (GCC 5.4.1) (judge verdict: Accepted)

//    https://www.youtube.com/watch?v=Dx6OQ7RvDXc

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef long long unsigned llu;
typedef pair<long long, long long> pll;
const long long inf = 2000000000000000000LL;	// 2e18
#define pi                  acos(-1.0)
#define fast                ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define shesh               "\n"
#define ff                  first
#define ss                  second
#define pb                  push_back
#define bp                  pop_back
#define pf                  push_front
#define fp                  pop_front
#define ub                  upper_bound
#define lb                  lower_bound
#define all(x)              x.begin(), x.end()
#define debug(x)            cout<<"debug "<<x<<"\n"
#define mest(a,b)           memset(a,b,sizeof(a))
int main()
{
    fast;
    ll t,n,m,i,j,k,l,temp,flag,len,shit,ans=inf;
    cin>>n;
    ll a[n];
    for(i=0;i<n;i++) cin>>a[i];
    for(i=1;i<=100;i++){
        for(j=0,temp=0;j<n;j++){
            temp+=((a[j]-i)*(a[j]-i));
        }
        ans=min(ans,temp);
    }
    cout<<ans;
}