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)

#include<iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <cmath>
#include <map>
#include <bitset>
#include <stack>
#include <queue>
#include <fstream>

using namespace std;
using ll=long long;

#define PRI(n) cout << n <<endl;
#define PRI2(n, m) cout << n << " " << m << endl;
#define PRI3(n, m, o) cout << n << " " << m<< " " << o << endl;

#define REP(i, n)  for(int i = 0; i < (int)n; ++i)
#define REPInv(i, n)  for(int i = n-1; i >= 0; --i)

#define FOR(i, t, n)  for(int i = t; i <= (int)n; ++i)
#define FORInv(i, t, n)  for(int i = t; i >= (int)n; --i)

const char alphabet[26] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                           's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
const ll MOD = 1e9 + 7;

ll GCD(ll a, ll b) {
    if (b == 0)return a;
    return GCD(b, a % b);
}

ll LCM(ll a, ll b) {
    ll gcd = GCD(a, b);
    return a / gcd * b;
}

ll nCr(int n, int r) {
    vector<ll> C(r + 1);
    C[0] = 1;
    FOR(i, 1, n) FORInv (j, min(i, r), 1)C[j] = (C[j] + C[j - 1]) % MOD;
    return C[r];
}

int N;
int A[110];

int main() {
    cin >> N;
    REP(i,N)cin >>A[i];
    ll ans = 1e18;
    FOR(i,-100,100){
        ll sum = 0;
        REP(j,N){
           sum += (i - A[j])*(i - A[j]);
        }
        if(ans>sum)ans = sum;
    }
    PRI(ans)
    return 0;
}