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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long 
#define llinf 1e17
#define nllinf -4e18
 
#define vll vector<ll>
#define sll set<ll>
#define pll pair<ll,ll> 
#define vpll vector<pair<ll,ll > >
 
#define repup(a,b,i) for(i=a;i<=b;i++)
#define repdown(a,b,i) for(i=b; i>=a; i--)
#define repupx(a,b,i,x) for(i=a;i<=b;i+=x)
	
#define repit(v,it) for(it=v.begin(); it!=v.end(); it++)
#define itall(v) v.begin(),v.end()
#define pb push_back
#define pf push_front
#define mp make_pair
#define fr first
#define sc second

#define err(n) cout<<n<<" ";
#define t1 cout<<",";

 ll MOD= 1e9 +7;

void add_self (ll& a, ll b){
	a+= b;
	if(a>=MOD)a-=MOD;
}

void sub_self (ll& a, ll b){
	a-=b;
	a= ((a%MOD)+MOD)%MOD;
}

ll dp[405][405];

int main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
ll i,j,k;
ll n;cin>>n;
ll a[n];repup(0,n-1,i)cin>>a[i];
ll b[n];
b[0]=a[0];
repup(1,n-1,i)b[i]=a[i]+b[i-1];

ll l,r;

repdown(0,n-1,l){
	repup(l,n-1,r){
		if(l==r)dp[l][r]=0;
		else{
			repup(l+1,r-1,k)	dp[l][r]= min(dp[l][r], dp[l][k]+dp[k][l]);
			
			dp[l][r]+= b[r];
			if(l!=0)dp[l][r]-=b[l-1];
		}
	}
}


cout<<dp[0][n-1];

}