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 fore(x,a,b) for(int x=a, qwerty=b; x<qwerty; x++)
#define ALL(a) (a).begin(),(a).end()
#define SZ(a) ((int)(a).size())
#define pb push_back
#define FIN ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define fst first
#define snd second
#define mset(a,v) memset((a),(v),sizeof(a))
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
using namespace std;
typedef long long ll;
const int N=200500;
int n;
int a[512];
ll dp[512][512],sp[512][512];
ll f(int i, int j){
}
int main(){FIN;
	cin>>n;
	fore(i,0,n)cin>>a[i];
	fore(i,0,n)dp[i][i]=0;
	fore(i,0,n)fore(j,i,n){
		ll s=0;fore(k,i,j+1)s+=a[k];
		sp[i][j]=s;
	}
	fore(l,1,n){
		fore(i,0,n){
			int j=min(i+l,n-1);
			if(i==j)continue;
			ll &r=dp[i][j];
			r=dp[i][i]+dp[i+1][j]+a[i]+sp[i+1][j];
			fore(k,i+1,j)r=min(r,dp[i][k]+dp[k+1][j]+sp[i][k]+sp[k+1][j]);
		}
	}
	cout<<dp[0][n-1]<<"\n";
}