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)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll n,K;
    scanf("%lld",&n);
    vector<ll> a(n);
    for(ll& x: a)
        scanf("%lld", &x);
    const ll MAX  = 1e18 ;   
    vector<vector<ll> > dp(n+1, vector<ll> (n+1,MAX) );
    vector<vector<ll> > sz(n+1, vector<ll> (n+1,0) );
    
    
    for(int l = 1; l<=n; l++)
    {
        for(int i =1; i<=n -l +1; i++)
        {
            int j = i + l-1;
            if(l==1)
                {
                    dp[i][j] = 0;
                    sz[i][j] = a[i-1];
                }
            else if (l==2)
                {
                    dp[i][j] = a[i-1] + a[j-1];
                    sz[i][j] = a[i-1] + a[j-1];
                }
            else 
                {
                    for(int x = i; x<j; x++)
                        {
                            dp[i][j] = min(dp[i][j], dp[i][x] + dp[x+1][j] +sz[i][x] + sz[x+1][j]);
                        }
                        sz[i][j] = sz[i][i] + sz[i+1][j];
                }
        }
    }
    
    cout<<dp[1][n]<<"\n";

}