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: Wrong Answer)

#include<bits/stdc++.h>

#define V vector<ll>
#define VV(v, n, m) vector<V> v(n, V(m));
#define _4(i, n) for(int i = 0; i < n; i++)
#define g_arr(arr, n) _4(i, n) cin >> arr[i];
#define p_vec(vec) _4(i, (int)vec.size()) cout << vec[i] << ' '; cout << endl;

typedef long long ll;

using namespace std;

int main(int argc, char * argv[]) {
    ios::sync_with_stdio(0);
    cin.tie(NULL);

    int n, idx;
    cin >> n;
    V v(n);
    g_arr(v, n);
    ll sum = 0, inter;

    _4(i, n - 1) {
        inter = 1e11;

        for(int j = 0; j < v.size() - 1; j++)
            if(v[j] + v[j + 1] < inter) {
                inter = v[j] + v[j + 1];
                idx = j;
            }

        if (idx + 1 < v.size()) {
            v[idx] += v[idx + 1];
            v.erase(v.begin() + idx + 1);
            sum += v[idx];
        }
    }

    cout << sum;

    return 0;
}