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>

#define INF 1000010000
#define nl '\n'
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define fi first
#define se second
#define pii pair<int,int>
#define pdd pair<double,double>
#define all(c) (c).begin(), (c).end()
#define SORT(c) sort(all(c))
#define sz(c) (c).size()
#define rep(i,n) for( int i = 0; i < n; ++i )
#define repi(i,n) for( int i = 1 ; i <= n; ++i )
#define repn(i,n) for( int i = n - 1 ; i >= 0 ; --i )
#define repf(j,i,n) for( int j = i ; j < n ; ++j )
#define die(s) {std::cout << s << nl;}
#define dier(s) {std::cout << s; return 0;}
#define dbg(var) {std::cout << #var << "  = " << var << nl;}
#define vi vector<int>
typedef long long ll;

using namespace std;


int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.precision(0); 

	int n;
	cin >> n;
	
	vector<ll> v(istream_iterator<int>{cin} , istream_iterator<int>());
	vector<ll> sum(n);
	partial_sum(all(v) , sum.begin());
	
	vector<vector<ll>> dp(n , vector<ll>(n , -1LL));
	function<ll(int,int)> rec = [&](int l , int r)
	{
		if(l == r)
			return 0LL;
			
		if(dp[l][r] != -1)
			return dp[l][r];
		
		ll mn = numeric_limits<ll>::max();
		for(int i = l ; i < r ; ++i)
			mn = min(mn , rec(l , i) + rec(i + 1 , r));

		return dp[l][r] = mn + sum[r] - (l > 0 ? sum[l - 1] : 0LL);
	};
	
	cout << rec(0 , n - 1);
	

	return 0;
}