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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)



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

public class Main {
	private static long sum(int L, int R, int[] a ) {
		long sum =0;
		for(int i=L; i<=R;i++) {
			sum+= a[i];
		}
		return sum;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in= new Scanner(System.in);
		int n = in.nextInt();
		int arr [] = new int[n];
		for( int i= 0; i<n; i++ )
			arr[i]= in.nextInt();
		long dp[][] = new long [n][n];
		for(int i=0;i<n; i++) {
			Arrays.fill(dp[i], (long)1e18 + 5);
			
		}
		
		for( int L= n-1; L>=0;L--) {
			for( int R=L;R<n;R++) {
				if(L==R)
					dp[L][R] = 0;
				else {
					long sum =sum(L,R,arr);
					for( int i=L; i<=R-1;i++) {
						dp[L][R] = Math.min(dp[L][R], dp[L][i] + dp[i+1][R] + sum);
					}
				}
			}
		}
		System.out.println(dp[0][n-1]);
	}

}