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>
#include <tuple>
using namespace std;

#ifdef LOCAL 
string to_string(char a) { return string(1,a); }
string to_string(bool a) { return a ? "1" : "0"; }
string to_string(const char* a) { return (string)a; }
string to_string(string a) { return a; }
string to_string(vector<bool> v) {
	string s = "["; for(unsigned i=0;i<v.size();i++){ s += ((i>0)?" ":"");
    s += char('0'+v[i]); } 	s += "]"; return s; }
template<class A, class B> string to_string(pair<A,B> p);
template<class T> string to_string(T v) {
	bool b = 1; string s = "["; for (const auto& x: v) {
		if (!b) {s += ", ";} b = 0; s += to_string(x); } s += "]"; return s; }
template<class A, class B> string to_string(pair<A,B> p) {
	return "("+to_string(p.first)+", "+to_string(p.second)+")"; }
void dbgf() { cerr << endl; }
template<class H, class... T> void dbgf(H h, T... t) {
	cerr << to_string(h); if (sizeof...(t)) cerr << ", ";
	dbgf(t...); }
#define dbg(...) cerr << #__VA_ARGS__ << " = ", dbgf(__VA_ARGS__)
#else
#define dbg(...) 0
#endif
#define boolout(x) cout<<(x?"YES":"NO")<<endl;
typedef long long ll;

const ll inf = 1e18 + 5;


void solve()
{
    int n;
    cin >> n;
    vector<ll> a(n);
    vector<ll> pref(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        if (i>0) {
            pref[i] = pref[i-1] + a[i];
        } else {
            pref[i] = a[i];
        }
    }
    dbg(pref);

    vector<vector<ll>> dp(n, vector<ll>(n, inf));

    for (int l = n-1; l >= 0; --l) {
        for (int r = l; r < n; ++r) {
            if (l==r) {
                dp[l][r] = 0;
            } else {
                for (int k = l; k < r; ++k) {
                    dp[l][r] = min(dp[l][r], dp[l][k] + dp[k+1][r] + pref[r] - (l==0?0:pref[l-1]));
                }
            }
        }
    }

    dbg(dp);
    cout << dp[0][n-1] << endl;

}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    //cin >> t; // read t. cin knows that t is an int, so it reads it as such.
    for (int i = 1; i <= t; ++i) {
        solve();
    }
    return 0;
}