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)

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define int long long
using namespace std;


int dp[405][405];
int pre[405];
signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);  
// #ifndef ONLINE_JUDGE
//    freopen ("INPUT.txt" , "r" , stdin);
//    freopen ("OUTPUT.txt" , "w" , stdout);
// #endif

    clock_t clk;
	clk = clock();

// --------------Code starts here---------------------------------------------------------------------

    int n;
    cin>>n;
    int a[n+1];
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        pre[i]=a[i]+pre[i-1];
    }

    fill_n(&dp[0][0],405*405,1e17);

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

        cout<<dp[1][n];

// -------------------------------------Code ends here------------------------------------------------------

    clk = clock() - clk;
	cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << "\n";
	return 0;
}