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>
using namespace std;
#define IO ios_base::sync_with_stdio(false)
#define ll long long
ll a[600000], vs[600000], dp[3000][3000], fs[500][500];
map<int,int>mp;
string s , s1;
ll f(ll l, ll r)
{
    ll i, j, n, mn=LLONG_MAX, k;
    if(l==r) return a[l];
    if(dp[l][r]) return fs[l][r];
    for(i=l ; i<r ; i++){
        k = f(l,i)+f(i+1,r);
        j = dp[l][i]+dp[i+1][r]+k;
        if(j<mn){
            mn = j ; fs[l][r] = k;
        }
    }
    dp[l][r]=mn; return fs[l][r];
}
int main()
{
    IO;
    ll j=0, n, m, k, m1, m2, a1=0, a2=0, l  = 1, t=3, mn, cnt=0, mx=0, h=0, mod=1e8+7, i; int tc=0;
    //cin>>t;
    //while(t--){
        cin>>n;

        for(i = 1; i<=n; i++)
              cin>>a[i];

        j = f(1, n); cout<<dp[1][n]<<endl;

    //}
    return 0;
}