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 int
ll dp[404][404];
ll arr[404];
ll solve(ll l,ll r){
	if(l>r)return 0;
	if(l==r)return arr[l];
	if(dp[l][r]!=-1)return dp[l][r];
	
//	cout<<l<<' '<<r<<'\n';
	
	ll x=arr[l]+min(solve(l+2,r),solve(l+1,r-1));
	ll y=arr[r]+min(solve(l,r-2),solve(l+1,r-1));
	dp[l][r]=max(x,y);
	return dp[l][r];
}

int main(){
	ll n,k;
	cin>>n;
//	memset(dp,-1,sizeof(dp));
	ll sum=0;
	for(ll i=0;i<n;i++){
		cin>>arr[i];
		sum+=arr[i];
		
		//cout<<"sun "<<i<<' '<<arr[i]<<' '<<sum<<'\n';
	}
	auto SUM=[&](ll l,ll r){
		
		ll sum=0;
		for(ll z=l;z<=r;z++){sum+=arr[z];}
		return sum;
	};
//	cout<<solve(0,n-1)<<' '<<sum<<'\n';
	for(ll l=0;l<=n-1;l++){
		for(ll i=0;i<n-l;i++){
			ll j=i+l;
			if(i==j){
				dp[i][j]=0;
			}
			else{
				//ll sum=0;
				dp[i][j]=1000000000;
				for(ll k=i+1;k<=j;k++){
					dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k][j]+SUM(i,j)
						
					);
					//sum+=arr[k];
				//	cout<<i<<' '<<j<<' '<<k<<' '<<dp[i][j]<<'\n';
				}
				
				//dp[i][j]+=sum;
			}
			
		}
	}
	
	
	cout<<dp[0][n-1]<<'\n';
	
}