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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mp make_pair
#define vl vector<ll>
#define vi vector<int>
#define sl set<ll>
#define sc set<char>
#define pi pair<int,int>
#define pl pair<ll,ll>
#define mpi map<ll,ll>
#define mpc map<char,ll>
#define forn(i,n) for(ll i=0;i<n;i++)
#define ff first
#define ss second
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define inf        1000000000000000005
#define MOD        1000000007

void solve(){
	ll n;
	cin>>n;
	ll a[n];
	forn(i,n){
		cin>>a[i];
	}
	//dp[i][j] denotes the cost in the interval i-j
	ll dp[n][n];
	for(ll l=n-1;l>=0;l--){
		for(ll r=l;r<n;r++){
			if(l==r){
				dp[l][r]=0;
			}
			else{
				dp[l][r]=inf;
				ll ans=0;
				for(ll i=l;i<=r;i++){
					ans+=a[i];
				}
				for(ll i=l;i<r;i++){
					dp[l][r]=min(dp[l][r],dp[l][i]+dp[i+1][r]+ans);
				}
			}
		}
	}
	cout<<dp[0][n-1];
}

int main(){
	ll t=1;
	//cin>>t;
	while(t--){
		solve();
		cout<<endl;
	}
}