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 ll long long 
#define m 1000000007
using namespace std;
ll dp[401][401];
ll p[401];
ll umin(ll a,ll b)
{
  if(a<b)
    return a;return b;
}
int main(int argc, char const *argv[])
{
  ll n;cin >> n;
  ll a[n];for(ll i=0;i<n;i++)
  {
  	cin >> a[i];
  	if(i==0)p[i]=a[i];
  	else p[i]=p[i-1]+a[i];
  	//cout << p[i] << " ";
  }
//cout << "\n"; 
  for(ll i=n-1;i>=0;i--)
  {
  	for(ll j=i;j<=n-1;j++)
  	{
  		if(i==j)dp[i][i]=0;
  		else
  		{
  			dp[i][j]=1e16;
  			for(ll k=i;k<=j-1;k++)
  			{	
  			ll dum = dp[i][k]+dp[k+1][j];
  			if(i==0)
  				dum+=p[j];
  			else
  				dum+=p[j]-p[i-1];
  			if(dp[i][j]>dum)
  				dp[i][j]=dum;
  			}
  		}
  	}
  }
  /*for(ll i=0;i<=n-1;i++)
  {
  	for(ll j=0;j<=n-1;j++)
  		cout << dp[i][j] << " ";
  	cout << "\n";
  }*/
  cout << dp[0][n-1];
  return 0;
}