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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("O3")

#define _GLIBCXX_DEBUG
#include <iostream>
#include <vector>
#include <string>
#include <string_view>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
using namespace std;

/*----------macros----------*/

//output================
#define endl '\n'
#define fcout(a) cout << fixed << setprecision((int)(a))

//typedef===============
typedef long long       ll;
typedef long double     ld;
typedef pair<int, int>  Pii;
typedef pair<ll, ll>    Pll;

//const number==========
constexpr ll INF9   = (ll)1e9 + 1;
constexpr ll INF18  = (ll)1e18 + 1;
constexpr ll MOD    = 1000000007;
constexpr ll MOD2   = 998244353;
constexpr ld PI     = 3.141592653589793L;

//member================
#define pb push_back
#define eb emplace_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second

//others================
#define all(x)          (x).begin(), (x).end()
#define rall(x)         (x).rbegin(), (x).rend()
#define rep(i, n)       for(int i = 0; (i) < (n); ++(i))
#define reps(i, n)      for(int i = 1; (i) <= (n); ++(i))
#define rrep(i, n)      for(int i = n - 1; (i) >= 0 ; --(i))
#define repbit(bit, n)  for(int bit = 0; (bit) < (1 << (n)); ++(bit))
#define ifbit(bit, i)   if((bit) & (1 << (i)))
#define CtoI(c)         int((c)) - int('0')
#define ItoC(c)         char((c) + '0')

//functions=============
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }

template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

//vector<int> dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};
//vector<int> dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };

/*--------additional--------*/



/*----------main------------*/

int main() {
/*----------fastio----------*/
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    std::ios::sync_with_stdio(false);
/*----------solve-----------*/

    int n; cin >> n;
    vector<int> a(n);
    int sum = 0;
    rep(i, n){
        cin >> a[i];
        sum += a[i];
    }

    int x = floorl(sum / n);
    int y = ceill(sum / n);
    int ans1 = 0, ans2 = 0;

    rep(i, n){
        ans1 += (a[i] - x) * (a[i] - x);
        ans2 += (a[i] - y) * (a[i] - y);
    }

    cout << min(ans1, ans2) << endl;

return 0;
}