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)

//In the name of Allah
//In the name of the compassionate of the merciful
//Ya Ali!
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename Type>
using ordered_multiset = tree<Type, null_type, less_equal<Type>, rb_tree_tag, tree_order_statistics_node_update> ;

typedef long long ll ;
typedef long double ld ;

const ll maxn = 410 ;
const ll INF = 3e18 ;

#define f first
#define s second
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define endll "\n"
#define sortt(v) sort(v.begin() , v.end())
#define reversee(v) reverse(v.begin() , v.end())

// for sqrt on interval and array , always sq = maxn/80 ;
// cout << *ss.find_by_order(index) ; -> return the value that indexed in index position
// cout << ss.order_of_key(value) ; -> lower_bound in ordered_set
//ifstream fin("input.txt");
//ofstream fout("output.txt");
// fflush(stdout);
ll mod = 1e9 + 7 , p1 = 999999929 , p2 = 999999937 ;

ll n , a[maxn] , dp[maxn][maxn] , per[maxn] ;

ll sum(ll l , ll r){
 return per[r]-per[l-1] ;
}

int main()
{
std::ios::sync_with_stdio(0) ;
cin.tie(0) ;
cout.tie(0) ;

cin >> n ;
for(int i = 1 ; i <= n ; i ++)cin >> a[i] ;

per[0] = 0 ;
for(int i = 1 ; i <= n ; i ++)per[i] = per[i-1]+a[i] ;

for(int r = 2 ; r <= n ; r ++){
  for(int l = r-1 ; l > 0 ; l --){
     dp[l][r] = INF;
     for(int k = l ; k < r ; k ++){
       dp[l][r] = min(dp[l][r] , dp[l][k]+dp[k+1][r]+sum(l,r)) ;
     }
  }
}
cout << dp[1][n] ;

 return 0;
}
/*
                  _     _   __
            /_\  (_    (_)  __)
           /   \  _) . (_) (__
*/