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>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
//using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int, int> pi;
typedef vector<int> vi;
template <class Key, class Compare = less<Key>>
using Tree = __gnu_pbds::tree<Key, __gnu_pbds::null_type, Compare, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;



int n;
ll a[404], p[404];
ll dp[404][404];

ll dfs(int i, int j) {
  ll& res = dp[i][j];
  if (res != -1) {
    return res;
  }
  if (i == 1 && j == n) {
    return res = 0;
  }
  res = 1e18;
  if (i > 1) {
    res = min(res, p[j] - p[i-1] + a[i-1] + dfs(i-1, j));
  }
  if (j < n) {
    res = min(res, p[j] - p[i-1] + a[j+1] + dfs(i, j+1));
  }
  return res;
}

void Main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  p[0] = 0;
  for (int i = 1; i <= n; ++i) {
    p[i] = p[i-1] + a[i];
  }
  memset(dp, -1, sizeof(dp));
  ll ans = 1e18;
  for (int i = 1; i <= n; ++i) {
    ans = min(ans, dfs(i, i));
  }
  cout << ans << '\n';
}

int main() {
  ios::sync_with_stdio(false);
//  cin.tie(nullptr);
  #ifdef _DEBUG
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
  #endif
  Main();
  return 0;
}