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;

#define el '\n'
#define ll long long

const int MAXN = 405;
const ll INF = 1e18L+5;

ll dp[MAXN][MAXN]; // dp[l][r] the min cost of combining l r into one vertex

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);

  int n;
  cin >> n;
  vector<ll> a(n);
  vector<ll> pref(n);
  for(int i = 0; i< n; i++){
    cin >> a[i];
    if(i==0) pref[i] = a[i];
    else pref[i] += pref[i-1]+a[i];
  }

  for(int i = 0; i < n; i++){
    for(int j = 0; j < n; j++){
      dp[i][j] = INF;
    }
  }

  for(int L = n-1; L >= 0; L--){
    for(int R = L; R < n; R++){
      if(L==R){
        dp[L][R] = 0;
      }else{
        for(int i = L; i < n; i++){
          dp[L][R] = min(dp[L][R], dp[L][i]+dp[i+1][R]+pref[R]-(L > 0 ? pref[L-1] : 0));
        }
      }
    }
  }

  cout << dp[0][n-1] << el;

  return 0;
}