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 vi vector<int>
#define pii pair<int, int>
#define vp vector<pii>
#define vs vector<string>
#define mii map<int, int>

void show(auto a){
  int i = 0;
  while(i < a.size()){
    cout<<a[i]<<" ";
    i++;
  }
  cout<<endl;
}

long long dp[410][410];

int main(){
  ios_base::sync_with_stdio(false);
  int n;
  cin>>n;
  vi a(n); for(int i =0; i < n; i++) cin>>a[i];

  for(int l = n; l >= 0; l--){
  	for(int r = 0; r <= n; r++){
  		if(l == r) dp[l][r] = 0;
  		else{
  			dp[l][r] = 1e18;
  			long long int s = 0;
  			for(int i = l; i <= r; i++){
  				s += a[i];
  			}
  			for(int i = l; i <= r; i++){
  				dp[l][r] = min(dp[l][r], dp[l][i] + dp[i+1][r] + s);
  			}
  		}
  	}
  }

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


  

}