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)

/*
created by sanyam singhal
*/

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

#define pb push_back
#define mp make_pair
#define ll long long int
#define mod 1000000007
#define w(t) int t;	cin>>t while(t--)
#define triplet pair<int,pair<int,int>>
#define vi vector<int>
#define vll vector<ll>
#define pi pair<int,int>
#define pll pair<ll,ll>
#define fastIO ios_base::sync_with_stdio(0);	cin.tie(0);	cout.tie(0);
#define loop(i, k, n) for(ll i=k; i<n; i+=1)
#define loop2(i, k, n) for(ll i=k; i>=n; i--)

const ll inf = 1e18;
const int N = 500;

ll dp[N][N];
ll sumij[N][N];

void take_fileIO(){
  #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  #endif
  return;
}

ll solve_by_dp(vector<ll> &a, ll i, ll j){
  if(j==i){
    return 0;
  }
  else if(dp[i][j]!=-1)
    return dp[i][j];
  else{
    ll ans = LLONG_MAX;
    for(ll k=i+1; k<=j; k++){
      ll part1 = solve_by_dp(a, i, k-1);
      ll part2 = solve_by_dp(a, k, j);
      ll curr_ans = part1 + part2 + sumij[i][k-1] + sumij[k][j];
      // cout<<curr_ans<<endl;
      if(curr_ans < ans){
        ans = curr_ans;
      }
    }
    return dp[i][j] = ans;
  }
}

void solve(){
  ll n; cin>>n;
  vector<ll>a(n);
  loop(i, 0, n) 
    cin>>a[i];

  memset(dp, -1, sizeof(dp));
  // loop(i, 0, n)
  //   cout<<a[i]<<" ";

  for(int i=0; i<n; i++){
    for(int j=i; j<n; j++){
      // for(int k=i; k<=j; k++){
      //   sumij[i][j] += a[k];
      // }
      if(i==j){
        sumij[i][j] = a[j];
        continue;
      }
      sumij[i][j] = a[j] + sumij[i][j-1];
    }
  }

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

  cout<<solve_by_dp(a, 0, n-1)<<endl;
  return;
}

int main(){

  fastIO;
  // take_fileIO();
  solve();
  return 0;
}