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>

using namespace std;

#define FOR(i,a,b)		for(int i=(a),_b=(b);i<(_b);++i)
#define FORD(i,a,b)		for(int i=(a),_b=(b);i>(_b);--i)
#define pb			push_back
#define mp			make_pair
#define	all(c)			(c).begin(),(c).end()
#define	present(c,x)		((c).find(x) != (c).end())
#define	cpresent(c,x)		(find(all(c),x) != (c).end())
#define	endl			'\n'

typedef long long		ll;
typedef unsigned long long	ull;
typedef unsigned char	 	byte;
typedef vector<int>		vi;
typedef pair<int, int>		pii;
typedef pair<ll, ll>		pll;
typedef vector<pii>		vpii;

const int MX = 405;

ll dp[MX][MX];

int main(int argc, char *argv[])
{
#ifdef	HTRINH_UNIT_TEST
	freopen(argv[1],"r",stdin);
	ifstream cin(argv[1]);
#endif
#if 0
	ofstream cout(argv[2]);
#endif
	ios :: sync_with_stdio(false);
	cin.tie(NULL);

	int n;
	cin >> n;
	ll a[n];
	FOR(i,0,n) cin >> a[i];
	ll sum[n];
	sum[0] = a[0];
	FOR(i,1,n) sum[i] = sum[i-1]+a[i];
	FOR(d,2,n+1) FOR(l,0,n) {
		int r = l+d-1;
		if (r >= n) break;
		ll tot = 0;
		if (l-1 >= 0) tot = sum[r]-sum[l-1];
		else tot = sum[r];
		ll best = 1LL<<60;
		FOR(k,l,r) best = min(best,dp[l][k]+dp[k+1][r]);
		dp[l][r] = best+tot;
	}
	cout << dp[0][n-1] << endl;
	return 0;
}