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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*--ILRS-- sr4saurabh  */
#include <bits/stdc++.h>
using namespace std; 
typedef long long int ll;
#define pb push_back 
#define popb pop_back
#define mp make_pair
#define ve vector
#define vii vector < int >
#define vll vector < ll > 
#define pii pair < int , int >
#define pll pair < ll , ll >
#define vpl vector < pll >
#define fi first
#define sz size 
#define len length 
#define se second 
//const ll mod=998244353;
const ll mod = 1000000000+7;
const ll N=10000000+6;
#define M_PI           3.14159265358979323846
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
   /** #ifndef ONLINE_JUDGE
       freopen ("INPUT.txt" , "r" , stdin);
       freopen ("OUTPUT.txt", "w" , stdout);
    #endif**/

       int n;
       cin>>n;
       ll arr[n];
       for(int i=0;i<n;i++)
        cin>>arr[i];

      ll dp[n+1][n+1];
      for(int l=n-1;l>=0;l--)
      {
        for(int r=l;r<n;r++)
        {
          dp[l][r]= 1e18l+5;
          ll sum=0;
          for(int j=l;j<=r;j++)
            sum=sum+arr[j];

          if(l==r)
            dp[l][r]=0;
          else
          {
            for(int i=l;i<=r-1;i++)
            {
              dp[l][r]=min(dp[l][r], dp[l][i]+dp[i+1][r]+sum);
            }
          }



        }
      }
      cout<<dp[0][n-1];


  














return 0;
}