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.*;
public class Main {
	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		long[][] dp=new long[n+1][n+1];
		int[][] s=new int[n+1][n+1];
		long[] a=new long[n+1];
		for (int i=1;i<=n;++i)
			for (int j=i;j<=n;++j)
				dp[i][j]=Long.MAX_VALUE;
		for (int i=1;i<=n;++i) {
			a[i]=in.nextLong()+a[i-1];
			dp[i][i]=0;
			s[i][i]=i;
		}
		for (int l=1;l<n;++l)
			for (int i=1;i+l<=n;++i) {
				int j=i+l;
				for (int k=Math.max(i,s[i][j-1]);k<=Math.min(j-1,s[i+1][j]);++k)
					if (dp[i][j]>dp[i][k]+dp[k+1][j]+a[j]-a[i-1]) {
						s[i][j]=k;
						dp[i][j]=dp[i][k]+dp[k+1][j]+a[j]-a[i-1];
					}
				}
		System.out.println(dp[1][n]);
	}
}