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: Wrong Answer)

    //Catchharsh
    #include<bits/stdc++.h>
    #define ll long long int 
    #define ld long double
    #define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
    #define F first
    #define S second
    #define pb push_back
    #define endl '\n'
    const ll MOD = 1e9+7;
    const ll INF = 1011111111;
    const ll LLINF = 1000111000111000111LL;
    const ld EPS = 1e-10;
    const ld PI = 3.14159265358979323;
    using namespace std;
    ll power (int x,int n)
    {
        if(n==0)
        return 1;
        else if(n%2==0)
        return power((x*x),n/2);
        else
        return x*power((x*x),(n-1)/2); 
    }
    vector<ll> seve;
    void  seveprime()
    {
      int prime[1000000]={0};
      for(int i=2;i*i<1000000;i++)
       {
        if(prime[i]==0)
         {
      	  for(int j=i*2;j<=1000000;j+=i)
           prime[j]=1;
         }
       }
       for(int i=2;i<=1000000;i++)
       {
       	if(prime[i]==0)
       		seve.pb(i);
       }
    }
    ll ans=0;
    vector<ll> v;
    void solve(ll n)
    { ll min=LLINF;
    	ll a,b,i;
    	for(i=0;i<n-1;i++)
    	{
           if(v[i]+v[i+1]<min)
           {
           	 min=v[i]+v[i+1];
           	 a=i;
           }
    	}
    	ans=ans+min;
    	v[a]=min;
    	v.erase(v.begin()+a+1);
    }
    int main()
    {   
         IOS;
         ll n,m,i,j,k;
         cin>>n;
         for(i=0;i<n;i++)
         {
         	cin>>j
         	;v.pb(j);
         }
         while(v.size()!=1)
         {
         	solve(v.size());
         }
         cout<<ans;
    }