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)

macro_rules !input {(source =$s :expr ,$($r :tt ) *) =>{let mut iter =$s .split_whitespace () ;let mut next =||{iter .next () .unwrap () } ;input_inner !{next ,$($r ) *} } ;($($r :tt ) *) =>{let stdin =std ::io ::stdin () ;let mut bytes =std ::io ::Read ::bytes (std ::io ::BufReader ::new (stdin .lock () ) ) ;let mut next =move ||->String {bytes .by_ref () .map (|r |r .unwrap () as char ) .skip_while (|c |c .is_whitespace () ) .take_while (|c |!c .is_whitespace () ) .collect () } ;input_inner !{next ,$($r ) *} } ;}
macro_rules !input_inner {($next :expr ) =>{} ;($next :expr ,) =>{} ;($next :expr ,$var :ident :$t :tt $($r :tt ) *) =>{let $var =read_value !($next ,$t ) ;input_inner !{$next $($r ) *} } ;}
macro_rules !read_value {($next :expr ,($($t :tt ) ,*) ) =>{($(read_value !($next ,$t ) ) ,*) } ;($next :expr ,[$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |read_value !($next ,$t ) ) .collect ::<Vec <_ >>() } ;($next :expr ,chars ) =>{read_value !($next ,String ) .chars () .collect ::<Vec <char >>() } ;($next :expr ,usize1 ) =>{read_value !($next ,usize ) -1 } ;($next :expr ,$t :ty ) =>{$next () .parse ::<$t >() .expect ("Parse error" ) } ;}

const INF: usize = std::usize::MAX;

fn main() {
   input! {
      n: usize,
      a: [usize; n]
   }
   let mut b = vec![0; n + 1];
   for i in 0..n {
      b[i + 1] = b[i] + a[i];
   }
   let mut dp = vec![vec![INF; n + 1]; n + 1];
   for i in 0..n {
      dp[i][i + 1] = 0;
   }

   for l in 2..n + 1 {
      for i in 0..n - l + 1 {
         for m in i + 1..i + l {
            dp[i][i + l] = std::cmp::min(dp[i][i + l], dp[i][m] + dp[m][i + l]);
         }
         dp[i][i + l] += b[i + l] - b[i];
      }
   }
   println!("{}", dp[0][n]);
}