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)

/***
**   AUTHOR::ASHUTOSH MOUDGIL
***/


#include<bits/stdc++.h>
using namespace std;

#define pb push_back
#define mp make_pair
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL);
#define F first
#define S second
#define int long long
//#define ll long long
#define all(x) x.begin(),x.end()
#define rsort(x) sort(all(x)); reverse(all(x));
#define endl '\n'
#define vi vector<int>
#define vlli vector<long long>
#define pii pair<int,int>
#define mod 1000000007
#define LMAX 1e18
#define deb(x) cout<<#x<<" : "<<x<<endl;
#define PI 3.141592654
#define deb1(x) cout << #x <<"="<< x <<endl;
#define deb2(x,y) cout << #x <<"="<<x<<", "<<#y<<"="<<y <<endl;
#define deb3(x,y,z) cout << #x <<"="<<x<<", "<<#y<<"="<<y <<", "<<#z<<"="<<z<<endl;
#define deb4(x,y,z,a) cout << #x <<"="<<x<<", "<<#y<<"="<<y <<", "<<#z<<"="<<z<< ", "<<#a <<"="<<a << endl;
#define debarr(x) cout << #x<<"::"<<endl;for(int i : x){cout << i <<" ";} cout <<endl;

int n;
int dp[500][500];


void mymain(){
    cin >> n;
    vi arr(n),pre(n,0);
    for(int i=0;i<n;i++)
        cin >> arr[i];
    pre[0] = arr[0];
    for(int i=1;i<n;i++)
        pre[i] = pre[i-1] + arr[i];
    for(int i=0;i<n;i++)
        dp[i][i] = 0;

    for(int l=1;l<=n;l++){
        for(int i=0;i+l<n;i++){
            int L = i, R = i+l;
            int upd=LMAX;
            for(int m = L;m<R;m++){
                upd = min(upd,dp[L][m]+dp[m+1][R]);
            }

                dp[L][R] = upd+pre[R];
                if(L!=0)
                    dp[L][R] -= pre[L-1];
        }
    }

    /*for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cout << dp[i][j] <<" ";
        }
        cout << endl;
    }*/

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


}

signed main(){
IOS;
///freopen("input.txt", "r", stdin);
///freopen("output.txt", "w", stdout);
int t=1;
//cin>>t;

for(int tt=0;tt<t;tt++){
   ///cout << "Case #" << tt+1 <<": ";
   mymain();
   }
return 0;
}
/*

*/