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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use text_io::*;

fn main() {
    const U_INF: usize = 1 << 60;
    const I_INF: isize = 1 << 60;
    let n: usize = read!();
    let mut a = vec![];
    for _ in 0..n {
        let ai: isize = read!();
        a.push(ai);
    }

    let mut sum_a = vec![0; n + 2];
    for i in 1..=n + 1 {
        sum_a[i] = sum_a[i - 1] + *a.get(i - 1).unwrap_or(&0);
    }


    let mut dp = vec![vec![None; n + 10]; n + 10];
    for i in 0..=n {
        dp[i][i] = Some(0);
    }
    println!("{}", solve(1, n, &mut dp, &sum_a));
}

fn solve(i: usize, j: usize, dp: &mut Vec<Vec<Option<isize>>>, a: &Vec<isize>) -> isize {
    match dp[i][j] {
        Some(t) => t,
        None => {
            let mut ans = 1 << 60;
            for k in i..j {
                ans = (ans).min(solve(i, k, dp, a) + solve(k + 1, j, dp, a) + a[j] - a[i - 1]);
            }
            dp[i][j] = Some(ans);
            ans
        }
    }
}