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: Wrong Answer)

use std::cmp::{max, min};
use std::error::Error;
use std::io;
use std::panic::resume_unwind;

pub fn main() {
    if let Err(e) = run() {
        panic!("failed.");
    }
}

fn run() -> Result<(), Box<Error>> {
    let mut input = String::new();
    io::stdin().read_line(&mut input);
    input.clear();
    io::stdin().read_line(&mut input);
    let slime: Vec<i64> = input.split_whitespace().map(|s| s.parse::<i64>().unwrap()).collect();
    let mut dp = vec![vec![std::i64::MAX; slime.len()]; slime.len()];
    let mut prefix = vec![0; slime.len() + 1];
    for i in 0..slime.len() {
        prefix[i + 1] = prefix[i] + slime[i];
    }
    do_solve(0, slime.len() - 1, &mut dp, &slime, &prefix);
    println!("{}", dp[0][slime.len() - 1]);
    Ok(())
}

fn do_solve(from: usize, to: usize, dp: &mut Vec<Vec<i64>>, slime: &Vec<i64>, prefix_sum: &Vec<i64>) -> i64 {
    if dp[from][to] != std::i64::MAX {
        return dp[from][to];
    }
    if from + 1 == to {
        return slime[from] + slime[to];
    }
    if from == to {
        return 0;
    }
    let mut result = std::i64::MAX;
    for i in from..to {
        let cost = do_solve(from, i, dp, slime, prefix_sum) + do_solve(i + 1, to, dp, slime, prefix_sum);
        let final_cost = prefix_sum[to + 1] - prefix_sum[from];
        result = min(result, cost + final_cost);
    }
    dp[from][to] = result;
    result
}