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>
using namespace std;

const int MAXN = 402;
long long int a[MAXN];
int n;

long long int find_ans(){
    long long int ans = 0;
    for(int i = 0; i < n - 1; ++i){
        long long int min_cost = 1e18, min_index = -1;
        for(int j = 0; j < n - 1 - i; ++j){
            if(a[j] + a[j + 1] < min_cost){
               min_index = j;
               min_cost = a[j] + a[j + 1];
            } 
        }
        
        a[min_index] = min_cost;
        for(int j = min_index + 1; j < n - 1 - i; ++j){
            a[j] = a[j + 1];
        }
        ans += min_cost;
    }
    return ans;
}

int main(){
	cin >> n;
	for(int i = 0; i < n; ++i){
      cin >> a[i];
    }
    
  	long long int ans = find_ans();
    cout << ans << "\n";
  	return 0;
}