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: Accepted)

#include<bits/stdc++.h>
#define FOR0(i,n) for(i=0;i<n;i++)
#define FOR(i,j,n) for(i=j;i<n;i++)
#define FORD(i,j,k) for(i=j;i>=k;i--)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define inf 1e17
#define ninf -1e17
#define endl '\n' 
#define she_told_me_dont_worry ios_base::sync_with_stdio (false) ; cin.tie(0) ; cout.tie(0) ;
// Use cout.flush() for interactive problems.
inline long long  MAX2(long long  a, long long int b){return (a)>(b)?(a):(b);}
inline long long  MAX3(long long  a, long long  b,long long  c){return (a)>(b)?((a)>(c)?(a):(c)):((b)>(c)?(b):(c));}
inline long long  MIN2(long long  a, long long  b){return (a)<(b)?(a):(b);}
inline long long  MIN3(long long  a, long long b,long long c){return (a)<(b)?((a)<(c)?(a):(c)):((b)<(c)?(b):(c));}
using namespace std;
typedef long long ll;
typedef pair<int,int> ii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector< ii > vii;
ll n;
ll a[405];
ll dp[405][405];
ll compute(int i, int j)
{
    if(i==j) return a[i];
    if(dp[i][j]!=-1) return dp[i][j];
    int k;
    ll ans = inf;
    ll ex = 0;
    FOR(k, i, j+1) ex += a[k];
    FOR(k, i, j)
    {
        ans = MIN2(ans, ex + compute(i, k) + compute(k+1, j));
    }
    return dp[i][j] = ans;
}
int main()
{
    she_told_me_dont_worry
    cin>>n;
    int i, j;
    FOR0(i, n+2)FOR0(j, n+2) dp[i][j] = -1;
    ll ans = 0;
    FOR0(i, n) 
    {
        cin>>a[i];
        ans -= a[i];
    }
    ans += compute(0, n-1);
    cout<<ans;
    cerr <<endl<< "Time elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '\n';
}