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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
ll mod=1e9+7;
#define endl '\n'
#define SIGMA 128
const ld PI=3.14159265358979;
#define fr(i,n) for(i=1;i<=n;i++)
#define rep(i,k,n) for(i=k;i<=n;i++)
#define w(x) int x;cin>>x;while(x--)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define ps(x,y) cout<<fixed<<setprecision(y)<<x<<endl
#define pb push_back
#define all(x) x.begin(),x.end()
#define vi vector<int>
#define vll vector<ll>
#define vvl vector<vll>
#define ppi pair<int,int>
#define ppl pair<ll,ll>
#define F first
#define S second 
#define LB lower_bound
#define UB upper_bound
#define mset multiset
#define uset unordered_set
#define umap unordered_map
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
ll binpow(ll x, ll y, ll p){  ll res = 1;x = x % p;if (x == 0)return 0;while (y > 0)  { if (y & 1) res = (res*x) % p; y = y>>1;x = (x*x) % p;}return res;}  
ll gcd(ll a, ll b){ if (b == 0) return a; return gcd(b, a % b); } 
bool parity(ll x,ll y){ bool f = ((x ^ y) < 0);return !f;}

ll dp[500][500];
int main()
{
    IOS

    ll n,i;
    cin>>n;
    vll v(n+1);
    vll pref(n+1,0);
    fr(i,n)
    {
        cin>>v[i];
        pref[i]=pref[i-1]+v[i];
    }

    fr(i,n)
    {
        for(int st=1;st+i-1<=n;st++)
        {
            int last=st+i-1;
            if(st==last)
            {
                dp[st][last]=0;
                continue;
            }
            dp[st][last]=mod*mod;
            for(int k=st;k<last;k++)
            {
                dp[st][last]=min(dp[st][last],dp[st][k]+dp[k+1][last]+pref[last]-pref[st-1]);
            }

        }
    }
    cout<<dp[1][n]<<endl;

    return 0;  
}