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 <iostream>
#include <algorithm>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <utility>
#include <string.h>
#include <map>
#include <stack>
#include <iomanip>
#include <chrono>
#include <random>
#include <math.h>
#include <time.h>
#include <assert.h>
#define rnd mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define ll long long
#define pb push_back 
#define mp make_pair
#define ff first
#define ss second
#define FIO ios_base::sync_with_stdio(false); cin.tie(NULL);
#define all(x) x.begin(),x.end() 
#define PII pair <ll,ll>
#define N 100005
#define MOD 1000000007
#define INF 1000000000000000000
using namespace std;

int main()
{
	ll n, A[405], sum[405], l, r;
	cin >> n;
	vector <vector<ll>> dp(401, vector<ll>(401, INF));
	sum[0] = 0;
	for (int i = 1; i <= n; ++i)
	{
		cin >> A[i];
		sum[i] = sum[i - 1] + A[i];
		dp[i][i] = 0;
	}
	for (int len = 2; len <= n; ++len)
	{
		for (int st = 1; st <= n - len + 1; ++st)
		{
			l = st;
			r = st + len - 1;
			for (int k = l; k < r; ++k)
			{
				dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r] + sum[r] - sum[l - 1]);
			}
		}
	}
	cout << dp[1][n];
}