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: Wrong Answer)

#include <bits/stdc++.h>

using LL = long long;

constexpr int MOD = 1'000'000'007;

int powmod (int a, int b, int MOD) {
	int ans = 1;
	while (b) {
		if (b & 1) ans = 1LL * ans * a % MOD;
		b >>= 1;
		a = 1LL * a * a % MOD;
	}
	return ans;
}


int main () {
	std::ios::sync_with_stdio (false);
	std::cin.tie (nullptr);

	int n; std::cin >> n;
	std::string s; std::cin >> s;

	std::vector <std::vector<LL>> dp;
	dp.assign (n+1, std::vector<LL> (n+1, 0));
	dp[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= i; ++j) {
			if (j > 0) dp[i][j] += 2 * dp[i-1][j-1];
			else dp[i][j] += dp[i-1][j];
			if (j < n) dp[i][j] += dp[i-1][j+1];
			dp[i][j] %= MOD;
		}
	}


	std::cerr << dp[n][(int)(s.length())] << "\n";

	std::cout << dp[n][(int)(s.length())] * powmod (powmod (2, MOD-2, MOD), (int)(s.length()), MOD) % MOD << "\n";

	return 0;
}