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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>

#define rep(i,n) for(int i = 0; i < (n); i++)
#define rrep(i,n) for(int i = (n)-1; i >= 0; i--)
#define rep1(i,n) for(int i = 1; i <= (n); i++)
#define rrep1(i,n) for(int i = (n); i > 0; i--)

#define ll long long
#define pi pair<int, int>
#define pll pair<ll, ll>

#define MOD 1000000007
#define INF 1000000000000000LL

using namespace std;

using namespace std;

#define MAXN 400

ll dp[MAXN+1][MAXN+1];

int main(){
  int n;cin>>n;
  vector<ll>a(n);
  rep(i, n)cin>>a[i];

  vector<ll> suma(n+1, 0);
  rep(i, n)suma[i+1]+=suma[i]+a[i];

  rep1(l, n)rep(i, n){
    int j = i+l;
    if(i>=j && j>=n)continue;
    dp[i][j]=INF;

    ll sum = suma[j+1]-suma[i];
    for(int m=i; m<=j; m++){
      dp[i][j] = min(dp[i][j], dp[i][m]+dp[m+1][j]+sum);
    }
//    cout<<i<<' '<<j<<' '<<dp[i][j]<<' '<<sum<<endl;
  }

  cout<<dp[0][n-1]<<endl;


  return 0;
  rep(i, n+1){
    rep(j, n+1)cout<<dp[i][j]<<' ';
    cout<<endl;
  }

  return 0;
}