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>
using namespace std;
int main() {
    #ifdef DEBUG
    std::ifstream in("/home/share/inputf.in");
    std::cin.rdbuf(in.rdbuf());
    #endif
    const int MAX_N = 400;
    const int64_t INF = 1e16;
    int N;
    cin >> N;
    // O(N^2)
    // (n, i) => cost
    // 左からi番目のスライムからn+1個が合体するときのコストとサイズ
    vector<int64_t> cost[MAX_N + 1];
    vector<int64_t> size[MAX_N + 1];
    for(int i = 1; i <= N; i++) {
        cost[i] = vector<int64_t>(N + 1 - i);
        size[i] = vector<int64_t>(N + 1 - i);
        for(int j = 0; j <= N - i; j++) {
            cost[i][j] = INF;
            size[i][j] = 0;
        }
    }
    for(int i = 0; i < N; i++) {
        cin >> size[1][i];
        cost[1][i] = 0;
    }
    for(int n = 2; n <= N; n++) {
        for(int i = 0; i <= N - n; i++) {
            for(int m = 1; m < n; m++) {
                size[n][i] = size[1][i] + size[n - 1][i + 1];
                cost[n][i] = min(cost[n][i], cost[m][i] + cost[n - m][i + m] + size[n][i]);
            }
        }
    }
    cout << cost[N][0] << endl;
    return 0;
}