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.ArrayList;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
	    Scanner sc = new Scanner(System.in);
	    int N = sc.nextInt();
	    long[] as = new long[N+1];
        long[][] dp = new long[N+1][N+1];
	    long[] prefixSum = new long[N+1];

	    for(int i=1; i<=N; ++i) {
	        as[i] = sc.nextLong();
	        prefixSum[i] = prefixSum[i-1] + as[i];
        }

//        d1(dp);

        for(int ij_d=1; ij_d<=N; ++ij_d) {
            for(int i=1; i+ij_d<=N; ++i) {
                int j = i+ij_d;
                dp[i][j] = Long.MAX_VALUE;
//                System.out.printf("%dx%d \n", i, j);
                for(int k=i; k<j; ++k) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k+1][j] + (prefixSum[j] - prefixSum[i-1]));
//                    System.out.printf("i,k %d,%d k+1,j %d,%d\n", i, k, k+1, j);
//                    System.out.printf("%d %d+%d\n", dp[i][j], dp[i][k], dp[k+1][j]);
                }
//                dp[i][j] += (prefixSum[j] - prefixSum[i-1]);
//                    System.out.printf("prefixSum %d,%d k+1,j %d,%d\n", i, k, k+1, j);

            }
        }

//        Util.<Long>d0(dp);
//        d1(dp);

	    System.out.println(dp[1][N]);
    }

    public static void d1(long[][] dp) {
        System.out.print("    ");

        for(int j=0; j<dp[0].length; ++j){
            System.out.printf("%3d ", j);
        }
        System.out.println();

        for(int i=0; i<dp.length; ++i){
            System.out.printf("%3d|", i);
            for(int j=0; j<dp[0].length; ++j){
                if(10000 < dp[i][j])
                    System.out.printf("%3d ", -1);
                else
                    System.out.printf("%3d ", dp[i][j]);
            }
            System.out.println();
        }
    }
}