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: Time Limit Exceeded)

#define DEBUG 1
#include <algorithm>
#include <cmath>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vll>;
using pll = pair<ll, ll>;
using tll = tuple<ll, ll, ll>;
#define all(v) (v).begin(), (v).end()
#define for1(i, n) for (ll i = 0; i < (n); i++)
#define for2(i, m, n) for (ll i = (m); i < (n); i++)
#define for3(i, m, n, d) for (ll i = (m); i < (n); i += (d))
#define rfor2(i, m, n) for (ll i = (m); i > (n); i--)
#define rfor3(i, m, n, d) for (ll i = (m); i > (n); i += (d))
#define INF 1111111111111111111LL
#define MOD 1000000007LL  // 10**9 + 7
#define print(...) print_1(__VA_ARGS__)
#define in(...) in_1(__VA_ARGS__)
#if DEBUG
#define dump(...) dump_1(#__VA_ARGS__, __VA_ARGS__)
#else
#define dump(...)
#endif
template <typename Head>
void dump_1(const char* str, Head&& h)
{
    cerr << str << ": " << h << '\n';
}
template <typename Head, typename... Tail>
void dump_1(const char* str, Head&& h, Tail&&... t)
{
    while (*str != ',') {
        cerr << *str++;
    }
    cerr << ": " << h << ' ';
    dump_1(str + 1, t...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& v)
{
    os << '(' << v.first << ", " << v.second << ')';
    return os;
}
template <typename T1, typename T2, typename T3>
ostream& operator<<(ostream& os, const tuple<T1, T2, T3>& v)
{
    os << '(' << get<0>(v) << ", " << get<1>(v) << ", " << get<2>(v) << ')';
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const map<T1, T2>& v)
{
    os << '{';
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ", ";
        }
        os << it->first << ':' << it->second;
    }
    os << '}';
    return os;
}
void Yes(void) { cout << "Yes\n"; }
void No(void) { cout << "No\n"; }
void YES(void) { cout << "YES\n"; }
void NO(void) { cout << "NO\n"; }
template <typename T>
bool chmax(T& a, const T& b)
{
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
bool chmin(T& a, const T& b)
{
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
void vin(vector<T>& v, ll len)
{
    for1 (i, len) {
        cin >> v[i];
    }
}
template <typename Head>
void in_1(Head& h)
{
    cin >> h;
}
template <typename Head, typename... Tail>
void in_1(Head& h, Tail&... t)
{
    cin >> h;
    in_1(t...);
}
template <typename Head>
void print_1(Head&& h)
{
    cout << h << '\n';
}
template <typename Head, typename... Tail>
void print_1(Head&& h, Tail&&... t)
{
    cout << h << ' ';
    print_1(t...);
}
//---------------------------------------------------------
ll memo[400][400];
ll A[400];
ll C[401];
ll calc(ll l, ll r)
{
    if (memo[l][r] != 0) {
        return memo[l][r];
    }
    if (l == r) {
        return 0;
    }
    ll ret = INF;
    for2 (i, l, r) {
        ll ret1 = calc(l, i) + calc(i + 1, r) + C[r + 1] - C[l];
        chmin(ret, ret1);
    }
    return ret;
}

void solve()
{
    ll N;
    in(N);
    for1 (i, N) {
        in(A[i]);
    }
    for1 (i, N) {
        C[i + 1] = C[i] + A[i];
    }
    print(calc(0, N - 1));
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(16);
    cerr << fixed << setprecision(16);
    solve();
}