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<iostream>
#include<bits/stdc++.h>
using namespace std;
int main(){
  int n;
  cin>>n;
list<long long int>a;
  for(int i=0;i<n;i++)
  {
    int x;
    cin>>x;
    a.push_back(x);
  }
  long long int ans=0;
  while(a.size()>1){
    auto it=a.begin();
    long long int mincost=*it;
    auto t1=a.begin();
    it++;
    mincost+=*it;
    auto t2=it;
    while(it!=prev(a.end())){
      long long int temp=*it;
      it++;
      temp+=*it;
      if(mincost>temp)
      {
        mincost=temp;
        t2=it;
        it--;
        t1=it;
        it++;
      }
    }
  /*  for(;it!=prev(a.end());it++){
      if(mincost>(*it)+(*(it+1)))
      {
        mincost=(*it)+(*(it+1));
        t1=it;
        t2=it+1;
      }
    }*/
    *t1=mincost;
    ans+=mincost;
    a.erase(t2); 
  }
  cout<<ans<<endl;

}