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: Wrong Answer)

#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define fi first
#define se second
#define pb push_back
#define vso(x) sort( x.begin() , x.end() )
#define aso(x) sort( x , x + x.size() )
#define print(x) cout << x << endl 
#define no cout << "NO\n" 
#define yes cout << "YES\n"
typedef long long ll ; 
typedef unsigned long long ull ;
typedef long double ld ;
typedef pair<ll, ll> pairs ;

// B6


int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    //freopen("problem_name.out", "w", stdout) ;
    //freopen("problem_name.in", "r", stdin) ;
    //ull h ; cin >> h ; srand(h) ;
    //int t ; cin >> t ; while(t--){}

    int n ; cin >> n ;
    vector <ull> v ;
    for ( int i = 0 ; i < n ; i++ )
    {
        ull a ; cin >> a ;
        v.pb(a) ;
    }
    ull sum = 0 ;
    for ( int j = 1 ; j < n ; j++ )
    {
        ull x = 0 , mini = 1000000000000000009 ;
        for ( int i = 1 ; i < v.size() ; i++ )
        {
            ull z = v[i-1] + v[i] ;
            if ( z <= mini )
            {
                mini = z ;
                x = i-1 ;
            }
        }
        sum += mini ;
        v.erase( v.begin() + x ) ;
        v.erase( v.begin() + x ) ;
        v.insert( v.begin() + x , mini ) ;
    }
    print(sum) ;
}