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: Wrong Answer)

#include<bits/stdc++.h>
#define fi first
#define se second
#define For(i,a,b) for(int i=a; i<=(b); ++i)
#define Fod(i,a,b) for(int i=a; i>=(b); --i)
#define pback push_back
#define speed ios::sync_with_stdio(false);
#define off_ cin.tie(0); cout.tie(0);
#define TLE 7e7 + 1
#define sz(x) (int) x.size()

using namespace std;
const int maxn = 1e3 + 5, maxm = 17, N = 1000;
const int mod = 99824453;
const long long oo  = 1e15 + 7;
typedef long long LL;
typedef pair<int, int> ii;

LL f[maxn][maxn];
int n, a[maxn];

int main()
{
    cin >> n;
    For(i, 1, n) cin >> a[i];
    For(i, 1, n)
        For(j, 1, n) f[i][j] = oo;
    For(i, 1, n) f[i][i] = a[i];
    For(len, 2, n)
        For(i, 1, n-len+1) {
            int j = i + len - 1;
            For(k, i, j-1) f[i][j] = min(f[i][k] + f[k+1][j], f[i][j]);
        }
    cout << f[1][n];
}