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>
#include<math.h>
#define N 100005
#define K 200005
#define MOD (long long )1e18L +5
#define ll long long int
using namespace std;
  ll a[405];
void add_self(int &a,int b)
{
    a+=b;
    if(a>=MOD)
    {
        a-=MOD;
    }
}
void sub_self(int &a,int b)
{
    a-=b;
    if(a<0)
    {
        a+=MOD;
    }
}
ll sum(ll l,ll r)
{
    ll sum1=0;
    for(ll i=l;i<=r;i++)
    {
        sum1+=a[i];
    }   
    return sum1;
}
ll min(ll a,ll b)
{
    return a<=b?a:b;
}
int main() {
    int n;
    cin>>n;
  
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    ll dp[n][n];
    // total cost of combining interval {i,j} into one vertex
    //think of it as a binary tree0
    for(int l=n-1;l>=0;l--)
    {
        for(int r=l;r<n;r++)
        {
            if(l==r)
            {
                dp[l][r]=0;
            }
            else{
            dp[l][r]=MOD;
            ll s=sum(l,r);
            for(int i=l;i<=r-1;i++)
            {
                dp[l][r]=min(dp[l][r],dp[l][i]+dp[i+1][r]+s);// we are calculating the value of the interval {l,r}
            }
            }
        }
            
    }
    cout<<dp[0][n-1]<<endl;

}