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>
using namespace std;
typedef long long ll;
typedef long l;


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

    // all the code goes here

    int n;cin>>n;
    
    ll dp[n];
    ll a[n];
    for(int i = 0; i < n; i++)
    {
    	cin>>a[i];
    	dp[i] = a[i];
    }
    ll sum  = 0;
    for(int i = 0; i < n-1; i++){
    	pair<int,int> a;
    	ll m = numeric_limits<ll>::max();
    	for(int j  = 0; j < n-i-1; j++){
    		if(dp[j]+dp[j+1] < m){
    			m = dp[j]+dp[j+1];
    			a = {j,j+1};
    		}
    	}

    	ll temp = dp[a.first]+dp[a.second];
    	dp[a.first] = temp;
    	for(int j  = a.second;j < n-i; j++)
    		dp[j] = dp[j+1];

    		// cout<<n-i-2<<"\n";
    	sum+=temp;
    }

    cout<<sum<<"\n";

    return 0;
}