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: Accepted)

#include<bits/stdc++.h>
#define M 1000000007
#define pb push_back
#define ll long long int
using namespace std;

int main()
{
	int n;
	cin>>n;
	int a[n];
	ll pref[n+1] = {0};
	int i,j;
	pref[0] = 0;
	for(i=0;i<n;i++)
	{
		cin>>a[i];
		pref[i+1] = pref[i]+a[i];
	}
	ll dp[n][n]={0};
	for(i=0;i<n;i++)
	{
		for(j=0;j<n-i;j++)
		{
			int k = j+i;
			if(j==k)
			{
				dp[j][k] = 0;
			}
			else
			{	
				int l = j;
				dp[j][k] = dp[j][l]+dp[l+1][k]+pref[k+1]-pref[j];
				for(l = j+1;l<k;l++)
				{
					dp[j][k] = min(dp[j][k],dp[j][l]+dp[l+1][k]+pref[k+1]-pref[j]);
				}
			}
		}
	}

	// for(i=0;i<n;i++)
	// {
	// 	for(j=0;j<n;j++)
	// 		cout<<dp[i][j]<<" ";
	// 	cout<<"\n";
	// }

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