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::vec::Vec;
use std::cmp::*;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
    let n: usize = read();
    let a: Vec<u64> = read_vec();
    // dp[i][j]はa_i ~ a_jを合成するときの最小コスト 答えはdp[0][n-1]
    // dp[i][j] = min(dp[i][l] + dp[l+1][j] + cost(i, l, j)) (i < l < j)
    // cost(i, l, j) = SUM(a_k) (for k = i ~ l) + SUM(a_k) (for k = l+1 ~ j)

    // コストを計算するための累積和
    let mut sum: Vec<u64> = vec![0; n+1];
    // dp_table
    let mut dp: Vec<Vec<u64>> = vec![vec![std::u64::MAX/2; n]; n];


    // 累積和を作成
    for i in 0..n {
      sum[i+1] = sum[i] + a[i];
    }

    // i == jの時、要素は１つなので合成できずコストは０、つまり、dp_tableの体格成分は０
    for i in 0..n {
      dp[i][i] = 0;
    }

    // dp_tableを更新
    // 区間の幅に関してループを回す
    for len in 1..n {
      for i in 0..n-len {
        // dp[i][i + len] = min(dp[i][i + len], )
        for j in i..i+len {
          dp[i][i + len]
            = min(dp[i][i + len],
                  dp[i][j] + dp[j+1][i + len]
                  + sum[i + len + 1] - sum[j+1]
                  + sum[j+1] - sum[i]);
        }
      }
    }
    println!("{}", dp[0][n-1]);
}