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>
#define REP(i, a, b) for(long long i=a; i<=b; ++i)
#define REPN(i, a, b) for(long long i=a; i<=b; --i)
#define rep(i, a, b) for(long long i=a; i<b; ++i)
#define rep0(i,b) for(long long i=0; i<b; ++i)
#define REP0(i,b) for(long long i=0; i<=b; ++i)
#include<cmath>
#include <iomanip>
using namespace std;


typedef long long ll;

int MODE = 1e9+7;

int main()
{

    int n;
    cin>>n;
    ll dp[n][n];
    memset(dp, -1, sizeof(dp));
    ll arr[n];
    rep0(i, n)
    {
        cin>>arr[i];
    }
    ll prefix[n+1];
    prefix[0] = 0;
    rep0(i, n+1)
    {
        prefix[i] = prefix[i-1] + arr[i];
    }
    for(int i=0; i<n; ++i)
    {
        dp[i][i] = 0;
    }

    for(int i=n-2; i>=0; --i)
    {

        for(int j=i+1; j<n; ++j)
        {
            dp[i][j] = 1e18+5;
            ll s =  prefix[j] - prefix[i-1];           
            for(int k = i; k<j; ++k)
            {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + s);
            }
        }
    }

    cout<<dp[0][n-1]<<endl;



    return 0;

}