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: Wrong Answer)

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

public class Main {

    static long rec(int i, int j, int[] b, long[][] dp) {
        if (dp[i][j] != Long.MAX_VALUE) {
            return dp[i][j];
        }

        if (j - i == 1) {
            return dp[i][j] = 0;
        }

        for (int k = 1; k < (j - i); k++) {
            dp[i][j] = Math.min(dp[i][j], rec(i, i + k, b, dp) + rec(i + k, j, b, dp));
        }

        return dp[i][j] = dp[i][j] + b[j] - b[i];
    }

    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int n = std.nextInt();
        int[] a = new int[n];
        int[] b = new int[n + 1];
        for (int i = 0; i < n; i++) {
            a[i] = std.nextInt();
        }

        for (int i = 0; i < n; i++) {
            b[i + 1] = a[i] + b[i];
        }

        long[][] dp = new long[n+1][n+1];
        for (int i = 0; i < n+1; i++) {
            Arrays.fill(dp[i], Long.MAX_VALUE);
        }

        long res = rec(0, n, b, dp);
        System.out.println(res);
    }
}