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>
using namespace std;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define sc(n) scanf("%d",&n)
#define scl(n) scanf("%lld",&n)
#define pr(n) printf("%d",n)
#define prl(n) printf("%lld",n)
#define nl printf("\n")
#define fr(i,n) for(i=0;i<n;i++)
#define rep(i,st,en) for(i=st;i<=en;i++)
#define repv(i,en,st) for(i=en;i>=st;i--)
#define fout cout<<fixed<<setprecision(7)
#define bi(n) __builtin_popcount(n)
#define bil(n) __builtin_popcountll(n)


typedef long long ll;
typedef pair<int,int> pii;
const int N = 410;
ll mod = 1e9+7;
ll fmod(ll b,ll exp){
    ll res =1;
    while(exp){if(exp&1ll)res=(res*b)%mod;
    b =(b*b)%mod;exp/=2ll;
    }
    return res;
}

ll A[N],pre[N];
ll dp[N][N];
ll INF = 1e18+5;
int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    int t=1,n,i,j,m;
    cin>>n;
    rep(i,1,n){
        cin>>A[i];
        pre[i]=pre[i-1];
        pre[i]+=A[i];
    }
    rep(i,1,n)rep(j,1,n)dp[i][j]=INF;

    rep(i,1,n)dp[i][i]=0;
    rep(i,1,n-1)dp[i][i+1]=A[i]+A[i+1];
    int sz=0;
    rep(sz,3,n)
    {
        rep(i,1,n)
        {
            int en = i+sz-1;
            if(en>n)break;

            for(int j=i+1;j<=en;j++)
            {
                dp[i][en]= min(dp[i][en],dp[i][j-1]+dp[j][en]+pre[en]-pre[i-1]);
            }

        }
    }
    cout<<dp[1][n]<<"\n";


    
    return 0;
}