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;
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
#define fastio() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define pb push_back
#define F first
#define S second
//cout<<fixed<<setprecision(11);
const string nl="\n";
const ll MOD=1e9+7;
const ll ARR_MAX = 405;
const ll VAL_MAX = 1e5+1;
const ll INF=1e14;

ll dp[ARR_MAX][ARR_MAX]; //dp[i][j] = minimum cost of construting [i...j]

void solve(){
    int n;cin>>n;
    ll a[n];
    ll s=0;
    ll cum[n+1];
    cum[0]=0;
    for(int i=0;i<n;i++){
    	cin>>a[i];
    	s+=a[i];
    	cum[i+1]=s;
    }

    for(int i=0;i<n;i++){
    	for(int j=0;j<n;j++){
    		dp[i][j]=INF;
    	}
    }
    for(int i=0;i<n;i++){
    	// dp[i][i]=a[i];
    	dp[i][i]=0;
    }
    for(int d=1;d<n;d++){
    	for(int i=0;i+d<n;i++){
    		for(int k=i;k<i+d;k++){
    			dp[i][i+d]=min(dp[i][i+d],dp[i][k] + dp[k+1][i+d] + cum[i+d+1]-cum[i]);
    		}
    	}
    }
    cout<<dp[0][n-1]<<nl;
}

int main(){
    fastio();
    int t=1;
    // cin>>t;
    while(t--){
        solve();
    }
    return 0;
}