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 bit(x,i) (x&(1<<i))  //select the bit of position i of x
#define IN(i,l,r) (l<i&&i<r) //the next for are for checking bound
#define LINR(i,l,r) (l<=i&&i<=r)
#define LIN(i,l,r) (l<=i&&i<r)
#define INR(i,l,r) (l<i&&i<=r)

#define F(i,L,R) for (int i = L; i < R; i++) //next four are for "for loops"
#define FE(i,L,R) for (int i = L; i <= R; i++)
#define FF(i,L,R) for (int i = L; i > R; i--)
#define FFE(i,L,R) for (int i = L; i >= R; i--)

#define getI(a) scanf("%d", &a) //next three are handy ways to get ints, it's also force you to use '&' sign
#define getII(a,b) scanf("%d%d", &a, &b)
#define getIII(a,b,c) scanf("%d%d%d", &a, &b, &c)

#define wez(n) int (n); scanf("%d",&(n)) //handy if the input is right after the definition of a variable
#define wez2(n,m) int (n),(m); scanf("%d %d",&(n),&(m))
#define wez3(n,m,k) int (n),(m),(k); scanf("%d %d %d",&(n),&(m),&(k))

#define whileZ int T; getI(T); while(T--)
#define lastEle(vec) vec[vec.size()-1]
#define SZ(x) ((int)((x).size()))

#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++) // traverse an STL data structure

#define ALL(c) (c).begin(),(c).end() //handy for function like "sort()"
#define ll long long //data types used often, but you don't want to type them time by time
#define ull unsigned long long
#define ui unsigned int

#define mp make_pair
#define fi first
#define se second

// for debug
#define printA(a,L,R) FE(i,L,R) cout << a[i] << (i==R?'\n':' ')
#define printV(a) printA(a,0,a.size()-1)

#define pb push_back
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;

ll a[401];
ll dp[401][401][2];     // value of slime, total cost incurred

ll best (int i, int j)  // return minimum cost
{
    if (dp[i][j][1]) return dp[i][j][1];
    if (i==j) {
        dp[i][j][0] = a[i];
        return dp[i][j][1] = 0LL;
    }
    dp[i][j][1] = 99999999999999LL;
    F(jj,i,j) {
        ll temp = best(i,jj) + best(jj+1,j);
        temp += dp[i][jj][0] + dp[jj+1][j][0];
        if (temp <= dp[i][j][1]) {
            dp[i][j][0] = dp[i][jj][0] + dp[jj+1][j][0];
            dp[i][j][1] = temp;    
        }
    }
    return dp[i][j][1];
}

int main()
{
    wez(n);
    F(i,0,n) cin >> a[i];
    cout << best (0,n-1) << "\n";
    return 0;
}