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<algorithm>
using namespace std;
int mod = 1000000007;
//long long dp[401][401] ;
vector<vector<long long>> dp(401, vector<long long>(401,1000000005));
long long a[401];
long long aa[401];
int main() {
  int n;
  cin >> n;
  
  for(int i = 1;i < n+1;i++){
    cin >> a[i];
    dp[i][i] = a[i];
    aa[i] = aa[i-1]+ a[i];
  }
  
  for(int i = 1;i < n;i++){
    dp[i][i+1] = a[i] + a[i+1];
  }
  
  long long res = 0;
  for(int i = 1;i < n; i++){
    long long tmp = 10000000000;
    for(int j = 1; j < n;j++){
      if(i+j > n){break;}
      for(int k = j; k < j+i;k++){
        dp[j][j+i] = min(dp[j][k]+dp[k+1][j+i],dp[j][j+i]);   
        //cout << j<< " " << k << " " << i+j << " " << dp[j][j+i] << endl; 
      }
      tmp = min(tmp,dp[j][j+i]);
    }
    res += tmp;
  }
  cout << res;
}