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>
using namespace std;
using ll = long long;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  ll N; cin >> N;
  vector<ll> A(N); for (ll i = 0; i < N; i++) cin >> A[i];
  vector<ll> ans = A;
  ll cost = 0;
  for (ll i = 0; i < N - 1; i++) {
    vector<ll> ans2 = ans;
    ll l = 0, temp = ans[0] + ans[1];
    for (ll j = 0; j < N - 1 - i; j++) {
      ll temp2 = ans[j] + ans[j + 1];
      if (temp2 < temp) {
        temp = temp2;
        l = j;
      }
    }
    ll p = 0;
    for (ll j = 0; j < N - 1 - i; j++) {
      if (p == l) {
        ans2[j] = ans[p] + ans[p + 1];
        cost += ans2[j];
        p++;
      } else {
        ans2[j] = ans[p];
      }
      p++;
    }
    // cout << ans[0]; for (int i = 1; i < ans.size(); i++) { cout << ' ' << ans[i]; }
    // cout << '\n';
    ans = ans2;
  }
  cout << cost << '\n';
  return 0;
}