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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

'use strict';
(function(input){
  input=input.trim().split(/\s+/).map(x=>x-0);
  const N=input.shift();
  const a=input;
  const dp=new Array(N).fill(0).map((x,i)=>new Array(N-i));
  //dp[i]はi回合成を行ったあとのslimeの配列。
  dp[0]=a;
  
  var ans=0;
  for(let i=1;i<N;i++){
    let m=Infinity;
    let J=-1;
    for(let j=0;j<N-i;j++){
      if(m>dp[i-1][j]+dp[i-1][j+1]){
        m=dp[i-1][j]+dp[i-1][j+1];
        J=j;
      }
    }
    ans+=m;
    
    for(let j=0;j<J;j++){
      dp[i][j]=dp[i-1][j];
    }
    dp[i][J]=m;
    for(let j=J+1;j<N-i-1;j++){
      dp[i][j]=dp[i-1][j+1];
    }
  }
  
  console.log(ans);
})(require('fs').readFileSync('/dev/stdin','utf8'));