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 (1.15.1) (judge verdict: Accepted)

use std::io::*;
const INF: u64 = 1u64 << 60;

fn rec(l: usize, r: usize, sum: &Vec<u64>, memo: &mut Vec<Vec<u64>>) -> u64 {
    if memo[l][r] != 0 {
        return memo[l][r];
    }
    if l == r {
        return 0;
    }
    let mut res = INF;
    for i in l..r {
        res = std::cmp::min(res, rec(l, i, &sum, memo) + rec(i + 1, r, &sum, memo));
    }
    res += sum[r + 1] - sum[l];
    memo[l][r] = res;
    res
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let a: Vec<u64> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();
    let mut sum: Vec<u64> = vec![0; n + 1];
    let mut memo: Vec<Vec<u64>> = vec![vec![0; n + 1]; n + 1];
    for i in 0..n {
        sum[i + 1] = sum[i] + a[i];
    }
    println!("{}", rec(0, n - 1, &sum, &mut memo))
}