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>
#include <ext/pb_ds/assoc_container.hpp>
#define ll long long int
#define inf 1000000000000
#define MOD 1000000007
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define for0(i, n) for (int i = 0; i < n; i++)
#define tc(t) int t; cin >> t; while (t--)
#define maxpq priority_queue<int>
#define minpq priority_queue<int, vector<int>, greater<int> >
#define vll vector<ll>
#define vi vector<int>
#define pb push_back
#define fi first
#define se second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mii map<int,int>
#define mll map<ll,ll>
#define S(x) (int)(x).size()
#define L(x) (int)(x).length()
#define ps(x,y)         fixed<<setprecision(y)<<x
using namespace std;
using namespace __gnu_pbds;

ll dp[405][405];
int n;
ll arr[405];
int main()
{
	IOS
/*#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif*/
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> arr[i];

	auto sum = [&](int l, int r)
	{
		ll s = 0;
		for (int i = l; i <= r; i++)
			s += arr[i];
		return s;
	};
	for (int l = n - 1; l >= 0; l--)
	{
		for (int r = l; r < n; r++)
		{
			if (l == r)
				dp[l][r] = 0;
			else
			{
				dp[l][r] = inf;
				ll s = sum(l, r);
				for (int i = l; i < r; i++)
				{
					dp[l][r] = min(dp[l][r], dp[l][i] + dp[i + 1][r] + s);
				}
			}
		}
	}
	cout << dp[0][n - 1];
}