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 int long long
#define double long double
#define SZ(x) ((int)(x).size())
#define rep(i,m,n) for(int i=(m);i<(n);++i)
#define fore(i,a) for(auto &i:a)
#define Yes cout << "Yes" << endl;
#define No cout << "No" << endl;
#define answer cout << ans << endl;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> using v = vector<T>;
template<class T> using vv = vector<v<T>>;
const int MOD=1e9+7;
const long long INF = 1LL << 60;
signed _main(); signed main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }

int rec(int l,int r,vv<int> &dp,v<int> &s){
  if(dp[l][r]!=-1) return dp[l][r];
  if(abs(l-r)<=1) return 0;
  int res=INF;
  for(int mid=l+1;mid<=r-1;mid++){
    chmin(res,rec(l,mid,dp,s)+rec(mid,r,dp,s)+s[r]-s[l]);
  }
  return dp[l][r]=res;
}

signed _main(){
  int N; cin >> N;
  v<int> w(N);
  rep(i,0,N) cin >> w[i];

  // 累積和
  vector<int> s(N+1, 0); // s[0] = 0 になる
  for (int i = 0; i < N; ++i) s[i+1] = s[i] + w[i];

  vv<int> dp(410,v<int>(410,-1));
  cout << rec(0,N,dp,s) << endl;
  return 0;
}