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)

//Radhe Radhe
#include<bits/stdc++.h>
//#include<boost/multiprecision/cpp_int.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//using namespace boost::multiprecision;
//#include<boost/math/common_factor.hpp>
#include <ext/numeric>
using namespace __gnu_cxx;
using namespace std;
#define int long long
#define endl "\n"
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define array(a,n) int a[n]; for(int i=0;i<n;i++) cin>>a[i];
#define loop(i,a,b,x) for(int i=a;i<b;i+=x)
const int MOD = 1000000007;

int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    int t = 1;
    //cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int  a[n];
        int pref[n + 1];
        pref[0] = 0;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            pref[i + 1] = pref[i] + a[i];
        }

        int dp[n][n];
        memset(dp, 0x3f, sizeof(dp));
        for (int i = 0; i < n; i++) {
            dp[i][i] = 0;
        }

        for (int l = 2; l <= n; l++) {
            for (int i = 0; i < n - l + 1; i++) {
                int j = i + l - 1;
                for (int k = i; k < j; k++) {
                    dp[i][j] = min(dp[i][j], (dp[i][k] + dp[k + 1][j]) + (pref[k + 1] - pref[i] + pref[j + 1] - pref[k + 1]));
                }
            }
        }

        cout << dp[0][n - 1] << endl;
    }
    return 0;

}