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)

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp> 
using namespace std;
//using namespace __gnu_pbds;
#define rep(i,a,b) for(i=(a);i<=(b);i++)
#define ll long long
#define pll pair<ll, ll>
#define pii pair<int,int>
#define pb push_back
#define F first
#define S second
#define mod 1000000007
#define maxn 100005
#define inf 1e18
#define boost ios::sync_with_stdio(false);cin.tie(0)
#define fr freopen("source.txt","r",stdin),freopen("output.txt","w",stdout)
#define SET(a,b) memset(a,b,sizeof(a))
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define all(x) x.begin(), x.end()
//template<typename T>
//using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

int main()
{
    boost;
    ll n,i,j,x,t,y,k;
    cin>>n;
    ll a[n];
    for(i=0;i<n;i++) cin>>a[i];
    ll ans=inf;
    for(i=1;i<=100;i++){
        ll temp=0;
        for(j=0;j<n;j++) temp+=(a[j]-i)*(a[j]-i);
        ans=min(ans,temp);
    }
    cout<<ans;
}