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)

/* 
    Reverse the problem: Imagine we have stick of length N and are told to cut it in the specified places
*/
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#include <iostream>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <stack>
#include <deque>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <utility>
#include <bitset>
 
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
 
#define LSOne(S) (S & (-S))
#define EPS 1e-9
 
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<double, double> pd;
typedef tree<pi, null_type, less<pi>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
 

int main()
{
    std::ios::sync_with_stdio(false);
    cin.tie(0); // BEFORE SUBMIT DON'T FORGET TO UNCOMMENT THIS
    int n;
    cin >> n;
    vector<int> vec(n+5);
    for(int i = 1; i <= n; ++i) cin >> vec[i];
    vector<ll> psum(n+5);
    for(int i = 1; i <= n; ++i) psum[i] = vec[i] + psum[i-1];

    vector<vector<ll>> dp(n+5, vector<ll>(n+5, 1e15L));
    // Base case
    for(int i = 1; i <= n; ++i) dp[i][i] = 0;

    for(int len = 1; len <= n; ++len)
    {
        for(int left = 1; left+len-1 <= n; ++left)
        {
            int right = left+len-1;
            for(int cut = left; cut < right; ++cut)
            {
                ll sumLeft = psum[cut] - psum[left-1];
                ll sumRight = psum[right] - psum[cut];
                dp[left][right] = min(dp[left][right], sumLeft + sumRight + dp[left][cut] + dp[cut+1][right]);
            }
        }
    }
    cout << dp[1][n] << "\n";
    return 0; 
}