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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
//#include <conio.h>
using namespace std;

typedef long long int ll;
typedef pair<ll,ll> pll;

#define rep(i,n)	for(ll i=0;i<(n);i++)
#define SZ(x) ((ll)x.size())
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define F first
#define S second
#define int ll
#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
// const ll MOD = INT_MAX;
const double PI = acos(-1);

void openfile(){
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
}
const int N = 405;
int dp[N][N];
int sum[N];

int func(int i,int j){
	if(dp[i][j])	return dp[i][j];
	if(i == j)		return dp[i][j] = 0;
	if(i == j-1)	return dp[i][j] = sum[j] - sum[i-1];
	int ans = 1e18+7;
	for(int k = i;k < j;k++){
		ans = min(ans,func(i,k) + func(k+1,j));
	}
	return dp[i][j] = ans + sum[j] - sum[i-1];
}


signed main(){
	IOS
	int n;	cin >> n;
	rep(i,n)	cin >> sum[i+1],sum[i+1] += sum[i];
	cout << func(1,n) << endl;

}