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;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
/*
    s.find_by_order(k) ---> returns iterator to the element at index(from 0) k
    s.order_of_key(val) ---> number of elements that are strictly smaller than the given value
*/
typedef long long int ll;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<int> vi;
typedef vector<long long int> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long int>> vvl;
typedef pair<int, int> pii;
typedef pair<long long int, long long int> pll;
typedef vector<pair<int, int>> vii;
typedef vector<pair<long long int, long long int>> vll;
#define fast() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define test() long long int tc; cin >> tc; cin.ignore(); while(tc--) 
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define mp make_pair
#define MOD 1000000007
#define SMOD 998244353
#define pq priority_queue
#define all(x) x.begin(),x.end()
#define SP(x) setprecision(x)
#define sp " "
#define inp(a) ll a;cin>>a;cin.ignore()
#define inps(a,b) ll a,b;cin>>a>>b;cin.ignore()
#define out(a) cout<<SP(20)<<a<<'\n'
#define outs(a,b) cout<<SP(20)<<a<<" "<<b<<'\n'
#define ins insert
#define er erase
#define lb lower_bound
#define ub upper_bound
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
#define inpa(a,n) ll a[n]; for(int i=0; i<n; i++){cin>>a[i];}
#define inpv(a,n) vl a; for(int i=0; i<n; i++){inp(x);a.pb(x);}
#define outa(a,n) for(int i=0;i<n;i++){cout<<a[i]<<" ";}cout<<'\n';
#define outv(v) for(auto i:v){cout<<i<<" ";}cout<<'\n';
set<char> vow = {'a','e','i','o','u','A','E','I','O','U'};
ll xpow(ll n, ll m){if(m==0){return 1;}else{ll x = xpow(n, m/2);if(m&1){return x*x*n;}else{return x*x;}}}
ll xmod(ll a, ll m){return (((a%m)+m)%m);}
ll xmodpow(ll n, ll m, ll y){if(m==0){return 1;}else{ll x = xmodpow(n, m/2, y);if(m&1){return xmod((xmod((xmod(x,y)*xmod(x,y)),y)*xmod(n,y)),y);}else{return xmod((xmod(x,y)*xmod(x,y)),y);}}}
 
int main() {
	fast();
 
    /*NA – JEONG – MO – SA – JI – MI – DA – CHAE – TZU*/
    
    ll n;
    cin >> n;
    inpv(a, n);
    ll ans = 1e18;
    for(int i=0; i<=100; i++){
        ll curr = 0;
        for(auto j:a){
            curr += ((j-i)*(j-i));
        }
        ans = min(ans, curr);
    }
    out(ans);
}