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 ll long long 
#define f(i, a, b) for(ll i=a;i<b;i++)
#define fi(i, a, b) for(ll i=b-1;i>=a;i--)
#define pb push_back
#define vi vector<int>
#define vl vector<ll>
#define ml map<ll,ll> 
#define pll pair<ll,ll>

ll pre[402];
ll dp[402][402];

ll ans(int l,int r)
{
   if(l==r)return 0;
   if(dp[l][r]!=-1)return dp[l][r];
   ll mini = (ll)(1e18);
   f(i,l,r)
   {
      mini = min(mini,ans(l,i)+ ans(i+1,r));
   }
   dp[l][r] = mini + pre[r]-pre[l-1];
   // cout<<l<<" "<<r<<" "<<dp[l][r]<<"\n";
   return dp[l][r];
}

void solve()
{
   ll n;
   cin>>n;
   ll a[n+1];
   f(i,1,n+1)cin>>a[i];
   pre[0]=0;
   f(i,1,n+1)pre[i] = pre[i-1] + a[i];
   f(i,1,n+1)
   {
      f(j,1,n+1) dp[i][j]=-1;
   }
   cout<<ans(1,n);

}

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