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)

/*
 * atcoder/dp_2nd/n.cpp
 */

// C++ 14
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring> // memset
#include <cassert>

using namespace std;

#define ll long long
#define MOD 1000000007

#define loop(__x, __start, __end) for(int __x = __start; __x < __end; __x++)
template <class T> bool chmin(T &a, T b) { if (a > b) {a = b; return true; } return false; }
template <class T> bool chmax(T &a, T b) { if (a < b) {a = b; return true; } return false; }

const int MAX = 420;
const ll INF = 100000000000000000;
ll N;
vector<ll> A;

// [l: r) までを合体させる最小コスト
ll dp[MAX][MAX];

void input() {
  cin >> N;
  ll a;
  loop(n,0,N) {
    cin >> a;
    A.push_back(a);
  }
}

// 累積和使わなくてもO(400^3)で間に合う
void solve() {
  for (int len=1; len<=N; len++) {
    for (int l=0; l+len<=N; l++) {
      if (len == 1) { dp[l][l+1] = 0; continue; }
      ll self = 0, child = INF;
      int r = l+len;
      for (int i=l; i<r; i++) self += A[i];
      for (int cut=l+1; cut<r; cut++) {
        chmin(child, dp[l][cut] + dp[cut][r]);
      }
      dp[l][r] = self + child;
    }
  }
  cout << dp[0][N] << endl;
}

void dump() {
  for (int len=1; len<=N; len++) {
    for (int l=0; l+len<=N; l++) {
      cout << "[" << l << ": " << (l+len-1) << "] dp = " << dp[l][l+len] << endl;
    }
  }
}

int main() {
  // cout.precision(15);
  input();
  solve();
  // dump();
  return 0;
}