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.io.*;
import java.util.*;
public class Main {
    public static void main(String args[]) {
        int n = 0;
        int k = 0;
        long MOD = 1000*1000*1000 + 7;
        int[] arr = null;
        long[] sums = null;
        try (Scanner scanner = new Scanner(System.in)) {
            n = scanner.nextInt();
            arr = new int[n];
            sums = new long[n];
            long sum = 0;
            for(int i = 0; i < n ;i++) {
                arr[i] = scanner.nextInt();
                sum += arr[i];
                sums[i] = sum;
            }
        }
        
        //dp[i][j] min cost for range i to j 
        long[][] dp = new long[n][n];
        for(int l = n - 1; l >= 0; l--) {
            for(int r = l; r < n; r++) {
                if(l == r)
                    dp[l][r] = 0;
                else {
                    dp[l][r] = Long.MAX_VALUE;
                    for(int i = l; i < r ;i++) {
                        dp[l][r] = Math.min(dp[l][r], dp[l][i] + dp[i+1][r] 
                                    + Math.abs((l-1 >=0?sums[l - 1]:0) - sums[r]));
                    }
                }
            }
        }
        
        // for(long[] d : dp)
        //     System.out.println(Arrays.toString(d));
        System.out.println(dp[0][n- 1]);
        
    }
    
}