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 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;

public class Main {


	public static void main(String[] args) {
		Main ans = new Main();
		System.out.println(ans.solve());
	}

	private long solve() {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int [] array = new int[N+1];

		for(int i=1;i<=N;i++){
			array[i] = sc.nextInt();
		}		 
		long [][] dp = new long[N+1][N+1];
 		long [][] sum = preProcess(array, N);
 		return solve(sum, 1, N, dp);
	}

	private long solve(long [][]sum, int i, int j, long[][] dp) {
		if (i==j) return 0;
		if (dp[i][j] != 0) return dp[i][j];

		long bestCost = (long) (1e12);
		for(int k=i;k<=j-1;k++) {
			long cost = sum[i][j] + solve(sum, i, k, dp) + solve(sum, k+1, j, dp);
			bestCost = Math.min(bestCost, cost);
		}
		dp[i][j] = bestCost;
		return bestCost;
	}

	private long [][] preProcess(int[]array, int N) {		 
		long [][] sum = new long[N+1][N+1];
		for(int i=1;i<=N;i++){
			for(int j=i;j<=N;j++){
				sum[i][j] = array[j] + ((i==j) ? 0 : sum[i][j-1]);
			}
		}
		return sum;
	}
}