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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#define ll long long int
#define FOR(i,a,b) for(int i=a;i<b;i++) 
using namespace std;

int main() {
	// your code goes here
	ll n;cin>>n;
	ll a[n];
	FOR(i,0,n)cin>>a[i];
	ll ans=0;
	ll sum=0;
	FOR(j,0,n-1){
		
		int i,f=0;
		ll x1=a[0],x2=a[1],p1,p2;
		
		for(i=0;i<n;i++){
			if(a[i]!=-1){
				if(f==0){
					x1=a[i];
					p1=i;
					f=1;
				}else{
					x2=a[i];
					p2=i;
					break;
				}
			}
		}
		
		for(i=p2+1;i<n;i++){
			if(a[i]==-1)continue;
			if(a[i]<x1){
				x1=x2;
				x2=a[i];
				p1=p2;
				p2=i;
			}
		}
		
		sum+=(x1+x2);
		a[p1]=x1+x2;
		a[p2]=-1;
		
	//	cout<<sum<<" "<<x1<<" "<<x2<<endl;
	//	for(int i=0;i<n;i++)cout<<a[i]<<" ";cout<<endl;
			
	}
	
	cout<<sum<<endl;
	return 0;
}