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)

#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 1000000000
#define INF 9223372036854775807
#define MOD 1000000007
#define mod 1000000007
#define INT_MAX_ 2147483647
#define EPS (1e-10)
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define rep(i,n)for (ll i = 0; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
#define me memset
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T> >;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp1(pair<ll,Pll> a,pair<ll,Pll> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se.fi>b.se.fi;
}
//----------------------------------------------------------------------

//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_t=clock();
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //-------------------------------
    ll n;cin>>n;
    V<ll>a(n);
    for(ll i=0;i<n;i++)cin>>a[i];
    ll cnt=n-1;
    ll ans=0;
    while(cnt--){
        n=a.size();
        V<ll>b;
        ll u1=-1,u2=-1;
        ll minm=INF;
        for(ll i=0;i<n;i++){
            chmin(minm,a[i]);
        }
        for(ll i=0;i<n;i++){
            if(a[i]==minm){
                if(i==0){
                    u1=0;
                    u2=1;
                }
                else if(i==n-1){
                    u1=n-1;
                    u2=n-2;
                }
                else{
                    u1=i;
                    if(a[i+1]>=a[i-1])u2=i-1;
                    else u2=i+1;
                }
                break;
            }
        }
        ans+=a[u1]+a[u2];
        a[u1]=a[u1]+a[u2];
        for(ll i=0;i<n;i++){
            if(i!=u2)b.pb(a[i]);
        }

        a.clear();
        for(ll i=0;i<b.size();i++){
            a.pb(b[i]);
        }
    }
    cout<<ans<<endl;
    //-------------------------------  
    //fclose(stdin);
    //fclose(stdout);
    //ll end_t=clock();cout<<"time="<<end_t-begin_t<<"ms"<<endl;
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------