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 <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <set>
#include <string>
#include <map>
#include <cmath>
#include <cstring>

#define fi first
#define se second
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define rrep(i,n) for (int i = 1; i < (n); ++i)
#define drep(i,n) for (int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define all(x) (x).begin(), (x).end()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define pb push_back
#define sz(x) (int)(x).size()
#define PQ(T) priority_queue<T, v(T), greater<T> >

using namespace std;
typedef long long ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;

const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;

const int mod = 1000000007;

const ll SMX = 405;

ll dp[SMX][SMX];
int main() {
    int n; cin >> n;
    vector<ll> a(n);
    rep(i,n) cin >> a[i];
    rep(i,SMX)rep(j,SMX) dp[i][j] = LINF;
    rep(i,n) dp[i][i] = 0;
    
    vector<ll> s(n+1, 0);
    rep(i,n) s[i+1] = s[i] + a[i];

    int sz = 1;
    while (sz < n) {
        rep(i,n) {
            if (i + sz >= n) break;
            rep(j,sz) {
                dp[i][i+sz] = min(dp[i][i+j]+dp[i+j+1][i+sz] + (s[i+j+1]-s[i]) + (s[i+sz+1]-s[i+j+1]), dp[i][i+sz]);
                //cout << "i: " << i << " j: " << (i+sz) << " : " << dp[i][i+sz] << endl;
            }

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

    return 0;
}