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;

template <typename T> T load() { T r; cin >> r; return r; }
template <typename T> vector<T> loadMany(int n) { vector<T> rs(n); generate(rs.begin(), rs.end(), &load<T>); return rs; }

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    auto n = load<int>();
    auto mass = loadMany<int>(n);
    auto joint = vector<vector<long long>>(n+1, vector<long long>(n+1, 0ll));
    for (auto l=1; l<=n; ++l)
        for (auto i=0; i+l<=n; ++i)
            joint[i][i+l] = mass[i] + joint[i+1][i+l];
    auto dyn = vector<vector<long long>>(n+1, vector<long long>(n+1, numeric_limits<long long>::max()));
    for (auto i=0; i<=n; ++i)
        dyn[i][i] = 0ll;
    for (auto i=0; i<n; ++i)
        dyn[i][i+1] = 0ll;
    for (auto l=2; l<=n; ++l)
        for (auto i=0; i+l<=n; ++i)
            for (auto m=i+1; m<i+l; ++m)
                dyn[i][i+l] = min(dyn[i][i+l], dyn[i][m] + dyn[m][i+l] + joint[i][m] + joint[m][i+l]);
    cout << dyn[0][n] << '\n';
}