Problem:
Score : 200 points

Problem Statement

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.

He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).

Find the minimum total cost to achieve his objective.

Constraints

1≦N≦100

-100≦a_i≦100

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum total cost to achieve Evi's objective.

Sample Input 1

2
4 8

Sample Output 1

8

Transforming the both into 6s will cost (4-6)^2+(8-6)^2=8 dollars, which is the minimum.

Sample Input 2

3
1 1 3

Sample Output 2

3

Transforming the all into 2s will cost (1-2)^2+(1-2)^2+(3-2)^2=3 dollars. Note that Evi has to pay (1-2)^2 dollar separately for transforming each of the two 1s.

Sample Input 3

3
4 2 5

Sample Output 3

5

Leaving the 4 as it is and transforming the 2 and the 5 into 4s will achieve the total cost of (2-4)^2+(5-4)^2=5 dollars, which is the minimum.

Sample Input 4

4
-100 -100 -100 -100

Sample Output 4

0

Without transforming anything, Evi's objective is already achieved. Thus, the necessary cost is 0.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES

#include <bits/stdc++.h>
#include <vector>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define int         long long
#define ll          long long
#define ld	    long double
#define pb          push_back
#define pp	    pop_back()
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define vii         vector<pii>
#define mi          map<ll,ll>
#define mii         map<pii,ll>
#define mp          make_pair
#define all(a)      (a).begin(),(a).end()
#define rall(a)     (a).rbegin(),a.rend()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define endl        '\n'
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define repj(j,a,b) for(ll j=a;j<b;j++)
#define lexistr(x,y)      lexicographical_compare(x.begin(),x.end(),y.begin(),y.end())
#define IOS	    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef tree<ll,null_type,less<ll>,rb_tree_tag, tree_order_statistics_node_update> indexed_set;

//CONSTANTS----------------------------------
#define PI 	    3.141592653589793238462643
const ll hell=1000000007;
const ll mod=1000000007;
//function-----------------------------------
ll GCD(ll a , ll b)
{
    if(b==0)
        return a;
    else
        return GCD(b,a%b);
}
ll LCM(ll a, ll b)
{
    return a*(b/GCD(a,b));
}
bool isPrime(ll n) {

    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (ll i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    return true;
}
ll fact(ll n)
{
    ll f=1;
    for(ll i=2;i<=n;i++)
	f=f*i;
    return f;
}
ll power(ll a,ll b)
{
  if(b==0)return 1;
  if(b==1)return a;
  ll t=power(a,b/2);
  t%=hell;
  t*=t;
  t%=hell;
  if(b%2)return (t*a)%hell;
  else
     return t;
}
//CODE------------------------------------------
ll solve()
{
    ll n; cin>>n;
    vi v(n);
    ll avg=0;
    rep(i,0,n)
        cin>>v[i];
    ll ans = INT_MAX;
    rep(i,-100,101)
    {
        ll sum = 0;
        repj(j,0,n) {
            sum += (v[j]-i)*(v[j]-i);
        }
        ans = min(ans, sum);
    }
    cout<<ans;
    return 0;
}
//MAIN------------------------------------------
signed main(){
    IOS
    #ifdef _DEBUG
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
    #endif
    int t=1;
    //cin>>t;
    while(t--) {
        solve();
        cout<<endl;
    }
    return 0;
}