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)

#include <bits/stdc++.h>
#define lli long long int
#define vec2 vector<vector<lli>> 
#define fio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define FRE freopen("explicit.in","r",stdin);freopen("explicit.out","w",stdout)
#define pll pair <lli,lli>
#define vll vector <lli>
#define vpll vector <pll>
#define f first
#define s second
#define pb push_back
#define priq priority_queue
#define unm unordered_map
#define pf push_front
#define minii LLONG_MAX
#define all(x) x.begin(), x.end()
#define mem(a,b) memset(a,b,sizeof(a))
#define ld double
#define eb emplace_back
const long double pi = 3.1415926535898;
const lli mod=1e9+7;
//const lli mod=100000000;
using namespace std;
lli lcm (lli a, lli b) {return ((a*b)/__gcd(a,b));}
lli modpower (lli a,lli b)
{
    if(b==0) return 1;
    lli c=modpower(a,b/2)%mod;
    if (b%2) return (((a*c)%mod)*c)%mod;
    else return (c*c)%mod;
}
lli power (lli a,lli b)
{
    if(b==0) return 1;
    lli c=power(a,b/2);
    if (b%2) return (a*c*c);
    else return c*c;
}
bool prime (lli k)
{
    if (k==1||k==0) return 0;
    bool flag=1;
    lli a=sqrt(k);
    for (int i=2;i<=a;i++)
    {
        if (k%i==0)
        {flag=0;break;}
    }
    return flag;
}
lli sieve (int n)
{
    bool qrr[n+1];
    for (int i=0;i<=n;i++)
    qrr[i]=1;qrr[0]=0;qrr[1]=0;
    for (int i=2;i*i<=n;i++)
    {
        if (qrr[i]==1)
        {
            for (int j=2*i;j<=n;j+=i)
            {qrr[j]=0;}
        }
    }
    return 0;
}
int main()
{
    fio;
    lli n;
    cin>>n;
    lli a[n];
    vll v;
    for (int i=0;i<n;i++)
    {
        cin>>a[i];
        v.pb(a[i]);
    }
    lli ans=0,mini,x;
    while (v.size()>1)
    {
        mini=1e18;x=0;
        auto it=v.begin();
        for (auto pt=v.begin();pt!=v.end();pt++)
        {
            auto pp=pt;
            pp++;
            if (pp==v.end())
            break;
            if (mini>*pt+*pp)
            {
                mini=*pt+*pp;
                it=pt;
            }
        }
        ans+=mini;
        *it=mini;
        it++;
        v.erase(it);
    }
    cout<<ans;
    return 0;
}