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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
//long long int a[410];
map<vector<long long int>, long long int>mp;

long long int sol(vector<long long int>v){
      if(v.size()==1)return 0;

      if(mp[v]!=0)return mp[v];

      long long int n=v.size(), ans=1000000000000;
      int j;

      for(int i=0; i<n; i++){
            if(i==0)j=n-1;
            else j=i-1;
            vector<long long int>v1=v;
            v1[i]+=v1[j];
            long long int sum=v1[i];
            v1.erase(v1.begin()+j);
            ans=min(ans, sol(v1)+sum);
      }
            /*for(int k=0; k<v.size(); k++)
                  cout<<v[k]<<' ';
                  cout<<'\n'<<ans<<"bgu"<<v.size()<<'\n';*/
      //cout<<ans;
      mp[v]=ans;
      return ans;
}

int main(){
      int n;
      cin>>n;
      vector<long long int>a(n);
      for(int i=0; i<n; i++)
            cin>>a[i];
      //cout<<mp[a];
      //cin>>n;
      cout<<sol(a);
}