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;
#include <unordered_map>
#define inputarr(a,n) for(ll i=0;i<n;i++) cin>>a[i];
#define prllarr(a,n) for(ll i=0;i<n;i++) cout<<a[i]<<" ";
#define pb push_back
#define ll long long
#define mod 1000000007
#define foi ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(0);
#define in(n) scanf("%lld",&n);
#define in2(x,y) scanf("%lld %lld",&(x),&(y));
#define in3(x,y,z) scanf("%lld %lld %lld",&(x),&(y),&(z));
#define out(n) printf("%lld\n",n);
#define out2(x,y) printf("%lld %lld\n",x,y);
#define test(t)  ll t; in(t);while(t--)
#define set(arr,n,s) for(ll i=0;i<n;i++){arr[i]=s;}


ll power(ll x,ll y,ll p){ll res=1;x=x%p;while(y>0){if(y&1){res=(res*x)%p;} y=y>>1;x=(x*x)%p;}return res;}
ll modInverse(ll a,ll p){return power(a,p-2,p);}//used with feemat little

ll gcd(ll x,ll y){if(x==0 || y==0){return max(y,x);}return gcd(y%x,x);}
ll gcdExtended(ll a,ll b,ll &x,ll &y){if(a==0){x=0;y=1;return b;}ll x1,y1;ll gcd = gcdExtended(b%a,a,x1,y1);x=y1-(b/a)*x1;y=x1;return gcd;}//o(log(b))
;
void prac(){
	#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
   #endif 
}
ll arr[403];
ll dp[405][405];
int main() 
{
  	// prac();
    ll n;
    in(n)
    for(ll i=1;i<=n;i++){
        in(arr[i])
    }
    arr[0]=0;
    for(ll i=1;i<=n;i++){
        arr[i]+=arr[i-1];
    }
 
    for(ll i=0;i<=400;i++){
        for(ll j=0;j<=400;j++){
            dp[i][j]=1e18;
        }
    }
    for(ll i=1;i<=n;i++){
        dp[i][i]=0;
    }
    for(ll len=2;len<=n;len++){
        for(ll i=1;i<=n;i++){
            ll end=i+len-1;
            if(end>n)break;
            for(ll k=i;k<end;k++){
                dp[i][end]=min(dp[i][k]+dp[k+1][end]+arr[end]-arr[i-1],dp[i][end]);
            }
        }
    }
    // for(ll i=0;i<=n;i++){
    //  for(ll j=0;j<=n;j++){
    //      cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
    //  }
    // }
    cout<<dp[1][n]<<endl;
    

   
}



/*error-----
convert every int to long long eg-1LL
create array with proper analysis of problem constrain
check mod also
*/