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: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long int

using namespace std;

int find(int a[],int n)
{
  ll c[n+1][n+1];
  ll sums[n+1][n+1];
  for(int i = 1; i<=n; i++)
  {
    int s = 0;
    for(int j = i; j<=n; j++)
    {
      s+=a[j];
      sums[i][j] = s;
    }
  }

  for(int s = 0; s<n; s++)
  {
    for(int i = 1,j = i+s; i<=n && j<=n; i++,j++)
    {
        if(i==j)
            c[i][j] = 0;
        else if(j==i+1)
            c[i][j] = sums[i][j];
        else
        {
            c[i][j] = INT32_MAX;
            for(int k = i; k<j; k++)
            {
                ll temp = c[i][k] + c[k+1][j] + sums[i][k]+sums[k+1][j];
                c[i][j] = min(c[i][j],temp);
            }
        }
    }
  }
  return c[1][n];
}

int main()
{
  int n;
  cin>>n;
  int a[n+1];
  for(int i = 1; i<=n; i++)
    cin>>a[i];
  cout<<find(a,n);
  return 0;
}