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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
// #include <ext/pb_ds/assoc_container.hpp> 
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds;   
#define ll long long
#define ull unsigned long long
// #define ordered_set tree<pair<ll, ll>, null_type, less<pair<ll, ll>>, rb_tree_tag, tree_order_statistics_node_update> 
ll mod=1e9+7;
#define PI 3.1415926535897932385
#define inf 9e18
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
typedef pair<long long, long long> ii;
#define S second
#define F first
ll max(ll a,ll b){if(a>b){return a;}return b;}
ll min(ll a,ll b){if(a<b){return a;}return b;}
#define MAXN 200005
// Comment this out for interactice problem
// #define endl '\n'
// string to integer stoi() Remember: it takes string not character
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
 
// To compile--> g++ -std=c++0x -o output one.cpp
// To run--> ./output

int main()
{ 
    fastio;
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);

    ll n;
    cin>>n;
    ll a[n+1];
    ll pref[n+1];
    pref[0]=0;
    for (int i = 1; i < n+1; ++i)
    {
        cin>>a[i];
        pref[i]=pref[i-1]+a[i];
    }

    ll dp[n+1][n+1];
    for (int i = 0; i < n+1; ++i)
    {
        for (int j = 0; j < n+1; ++j)
        {
            dp[i][j]=inf;
        }
    }

    for (int i = 1; i < n+1; ++i)
    {
        dp[i][i]=0;
    }
    for (int i = 1; i < n; ++i)
    {
        dp[i][i+1]=a[i]+a[i+1];
    }

    for (int l = 2; l<=n; ++l)
    {
        for (int i = 1; i < n-l+1; ++i)
        {
            ll j = i+l;
            for (int k = i; k<j; ++k)
            {
                dp[i][j] = min(dp[i][j], dp[i][k]+dp[k+1][j]);
            }
            dp[i][j]+=pref[j]-pref[i-1];
        }
    }

    cout<<dp[1][n];

    return 0;
}