Problem:
Score : 100 points

Problem Statement

There are N slimes lining up in a row.
Initially, the i-th slime from the left has a size of a_i.

Taro is trying to combine all the slimes into a larger slime.
He will perform the following operation repeatedly until there is only one slime:

Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.

Find the minimum possible total cost incurred.

Constraints

All values in input are integers.

2 \leq N \leq 400

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_N

Output

Print the minimum possible total cost incurred.

Sample Input 1

4
10 20 30 40

Sample Output 1

190

Taro should do as follows (slimes being combined are shown in bold):

(10, 20, 30, 40) → (30, 30, 40)

(30, 30, 40) → (60, 40)

(60, 40) → (100)

Sample Input 2

5
10 10 10 10 10

Sample Output 2

120

Taro should do, for example, as follows:

(10, 10, 10, 10, 10) → (20, 10, 10, 10)

(20, 10, 10, 10) → (20, 20, 10)

(20, 20, 10) → (20, 30)

(20, 30) → (50)

Sample Input 3

3
1000000000 1000000000 1000000000

Sample Output 3

5000000000

The answer may not fit into a 32-bit integer type.

Sample Input 4

6
7 6 8 6 1 1

Sample Output 4

68

Taro should do, for example, as follows:

(7, 6, 8, 6, 1, 1) → (7, 6, 8, 6, 2)

(7, 6, 8, 6, 2) → (7, 6, 8, 8)

(7, 6, 8, 8) → (13, 8, 8)

(13, 8, 8) → (13, 16)

(13, 16) → (29)

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	static Scanner sc = new Scanner(System.in);
	static long MOD = 1_000_000_007;

	public static void main(String[] args) {
		int N = sc.nextInt();
		long[] sum = new long[N + 1];
		int[] A = new int[N];
		for (int i = 0; i < N; i++) {
			A[i] = sc.nextInt();
			sum[i + 1] = sum[i] + A[i];
		}
		long[][] dp = new long[N + 1][N + 1];
		for (long[] a : dp) {
			Arrays.fill(a, Long.MAX_VALUE / 8);
		}
		for (int i = 0; i < N; i++) {
			dp[i][i + 1] = A[i];
		}
		for (int i = 2; i <= N; i++) {
			for (int j = 0; j + i <= N; j++) {
				for (int k = 1; k < i; k++) {
					dp[j][j + i] = Math.min(dp[j][j + i], dp[j][j + k] + dp[j + k][j + i] + sum[j + i] - sum[j]);
				}
			}
		}
		System.out.println(dp[0][N] - sum[N]);
	}
}