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.Scanner;

public class Main {

	static long[][][] dp;
	static long[] largeArray;

	public static void main(String[] args) {

		Scanner scan = new Scanner(System.in);

		int n = scan.nextInt();

		largeArray = new long[n];

		for (int i = 0; i < n; i++) {

			largeArray[i] = scan.nextLong();
		}

		dp = new long[n + 1][n + 1][2];

		for (int end = 1; end <= n; end++) {

			for (int start = n; start > 0; start--) {

				dpMethod(start, end);

			}
		}

		println(dp[1][n][1]);
	}

	public static void dpMethod(int start, int end) {

		if (start > end) {

			return;

		} else if (start == end) {

			dp[start][end][0] = largeArray[start - 1];

		} else {

			long costResult = Long.MAX_VALUE;

			for (int i = 1; i <= end - start; i++) {

				long target = dp[start + i][end][0]
						+ dp[start + i][end][1]
						+ dp[start][start + i - 1][0]
						+ dp[start][start + i - 1][1];

				costResult = Math.min(costResult, target);
			}

			long resultLarge = dp[start + 1][end][0] + dp[start][start][0];

			dp[start][end][0] = resultLarge;
			dp[start][end][1] = costResult;
		}
	}

	public static void println(Object target) {
		System.out.println(target);
	}
}