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<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#include <bitset>
#include <cmath>
#include <limits>
#include <iostream>
#include<map>
#include<tuple>
using namespace std;
#define INF 1100000000000
#define MAX 1100000
#define MOD 1000000007
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<pair<int,int>,int> p;
typedef pair< pair<int,int>, int> p; 
#define bit(n,k) ((n>>k)&1) /*nのk bit目*/
#define rad_to_deg(rad) (((rad)/2/M_PI)*360)
//__builtin_popcount(S)

//https://kyopro-friends.hatenablog.com/entry/2019/01/12/231035

bool flag[410][410];
ll S[410],a[410];
ll dp[410][410];

ll f(int l,int r){
    if(flag[l][r]) return dp[l][r];
    flag[l][r]=true;
    if(l==r) return 0;
    ll fans=INF;
    for(int m=l;m<r;m++){
        fans=min(fans,f(l,m)+f(m+1,r));
    }
    return dp[l][r]=fans+S[r]-S[l-1];
}

int main(){
    for(int i=0;i<410;i++){
        for(int j=0;j<410;j++) flag[i][j]=false;
    }
    int N; cin>>N;
    for(int i=1;i<=N;i++) cin>>a[i];
    for(int i=1;i<=N;i++){
        S[i]=S[i-1]+a[i];
    }
    cout<<f(1,N)<<endl;
}