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: Wrong Answer)

/* ****AWARE**** */
#include<bits/stdc++.h>
#include <limits.h> 
#define cut ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ll          long long
#define pb          push_back
#define endl        '\n'
#define pii         pair< ll, ll>
#define vi          vector< ll>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define rep(i,a,b)  for(  ll i=a;i<b;i++)
#define lbnd        lower_bound
#define ubnd        upper_bound
#define bs          binary_search
#define mp13       make_pair
using namespace std; 
#define MAX  3000
#define MODULO 2
using namespace std;   
#define hell 1000000007 
#define hell1 30000000000000000
#define time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
#define mysizeof(i) (char*)(&i+1)-(char *)(&i);
ll T;
//double dp[MAX][MAX];
vector<ll>v;
void solve()
{    ll i,j,n,a,k,mn,cnt=1,ans,sum=0;
     cin>>n;
     for(i=0;i<n;i++)
     	cin>>a,v.pb(a);

     while(n-cnt!=0)
     {
     mn=1e13;
     for(i=0;i<(n-cnt);i++)
     {
       if(v[i]+v[i+1]<mn)
       {
       	ans=i;
       	mn=v[i]+v[i+1];
       }
     }
     sum+=(v[ans]+v[ans+1]);
     v[ans]=v[ans]+v[ans+1];
     cnt++;
     for(i=ans+1;i<n;i++)
     {
     	v[i]=v[i+1];
     }
    }
    cout<<sum;
    
}
           
int main(){ 
cut; 
ll t; 
//cin>>t;
//T=t;
 t=1;
while(t--)
solve();
return 0;
}