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 <iostream>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <fstream>
#include <sstream>
#include <iomanip>

using namespace std;

long long INF = 1000000000000000000;

void cout_vec(vector<long long> &vec){
    for(int i=0; i<vec.size(); i++){
        if (i!=0){
            cout << ' ';
        }
        cout << vec[i];
    }
    cout << endl;
}

void cout_mat(vector<vector<long long>> &mat){
    for(int i=0; i<mat.size(); i++){
        vector<long long> vec = mat[i];
        cout_vec(mat[i]);
    }
}

int main(){
    int N;
    cin >> N;
    vector<long long> a(N+1,0);
    for(int i=0; i<N; i++){
        int b;
        cin >> b;
        a[i+1]=a[i]+b;
    }
    vector<vector<long long> > dp(N+1,vector<long long>(N+1,INF));
    for(int n=1; n<=N; n++){
        for(int i=0; i<=N-n; i++){
            if(n==1)dp[i][i+n]=0;
            for(int k=1; k<n; k++){
                dp[i][i+n]=min(dp[i][i+n], dp[i][i+k]+dp[i+k][i+n]+a[i+n]-a[i]);
            }
        }
    }
    //cout_mat(dp);
    cout << dp[0][N] << endl;
}