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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.io.*;

public class Main 
{
	public static void main(String[] args) { new Main(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);
	
	int n, c;
	long mod = (long)1e9 + 7;
       	int[] a, b;
	long[][] pre, dp;
	
	Main() {
		n = in.nextInt();
		c = in.nextInt();
		a = new int[n]; b = new int[n];
		for (int i = 0; i < n; a[i++] = in.nextInt());
		for (int i = 0; i < n; b[i++] = in.nextInt());

		pre = new long[n][c + 1];
		for (int i = 0; i < n; i++)
			for (int j = 0; j <= c; j++)
				for (int k = a[i]; k <= b[i]; k++) {
					pre[i][j] += pow(k, j);
					pre[i][j] %= mod;
				}

		dp = new long[n][c + 1];
		out.println(dp(0, c));
		out.close();
	}
	
	long dp(int idx, int left) {
		if (left == 0)
			return 1;
		if (idx == n)
			return 0;
		if (dp[idx][left] != -1)
			return dp[idx][left];

		long ans = 0;
		for (int i = 0; i <= left; i++) {
			ans += mult(pre[idx][i], dp(idx + 1, left - i));
			ans %= mod;
		}
		return dp[idx][left] = ans;
	}

	long mult(long a, long b) {
		return (a % mod)*(b % mod) % mod;
	}

	long pow(long a, long b) {
		long res = 1L;
		while (b > 0) {
			res *= res;
			if ((b & 1) > 0)
				res *= a;
			res %= mod;
			b >>= 1;
		}
		return res;
	}
	
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}