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: Wrong Answer)


#include "bits/stdc++.h"
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;

#define ub              upper_bound
#define lb              lower_bound
#define isrt            insert
#define clr             clear
#define rsz             resize
#define ff              first
#define ss              second
#define lli             long long int
#define pb              push_back
#define pf              push_front
#define mkp              make_pair
#define pii             pair<int,int>
#define vi              vector<int>
#define mii             map<lli,lli>
#define pqb             priority_queue<int>
#define pqs             priority_queue<int,vi,greater<int> >
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define mod             1000000007
#define INF             1e9
#define ps(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(x)            int x; cin>>x; while(x--)
#define all(v)          v.begin(),v.end()
#define endl            "\n"


typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;

lli maxm(lli a, lli b)
{
  return (a >= b ? a : b);
}


lli minm(lli a, lli b)
{
  return (a <= b ? a : b);
}

lli power(lli x, lli y, lli p)
{
  lli res = 1;

  x = x % p;
  if (x == 0)
    return 0;

  while (y > 0)
  {

    if (y & 1)
      res = (res * x) % p;


    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}

lli nCrmodp(lli n, lli r, lli p)
{
  if (r == 0)
    return 1;
  if (r == 1)
    return n;


  lli pro = 1, k = n;
  while (k > r)
  {
    pro *= k;
    pro %= p;
    --k;
  }

  k = 2;
  while (k <= (n - r))
  {
    pro *= power(k, (p - 2), p);
    pro %= p;
    ++k;
  }

  return pro;
}


string num_str(lli n, lli k)
{
  string str = "";
  lli x = n;
  while (x != 0)
  {
    string s = to_string(x % k);
    str = s + str;
    x /= k;
  }

  return str;
}

void c_p_c()
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
}

int main()
{
  c_p_c();

  lli n, k;
  cin >> n;

  lli sum = 0;
  vector<lli> ev, odd;

  for (lli i = 0; i < n; ++i)
  {
    cin >> k;
    ev.pb(k);
  }

  for (lli i = 1; i < n; ++i)
  {
    lli br, minim = 1e17;

    if (i % 2 != 0)
    {
      for (lli j = 0; j + 1 < ev.size(); ++j)
      {
        if ((ev[j] + ev[j + 1]) < minim)
        {
          minim = (ev[j] + ev[j + 1]);
          br = j;
        }
      }

      odd.clr();
      ev[br] = minim;
      br++;
      sum += minim;
      //cout << minim << endl;
      for (lli j = 0; j < ev.size(); ++j)
      {
        if (j != br)
          odd.pb(ev[j]);
      }
    }

    else
    {

      for (lli j = 0; j + 1 < odd.size(); ++j)
      {
        if ((odd[j] + odd[j + 1]) < minim)
        {
          minim = (odd[j] + odd[j + 1]);
          br = j;
        }
      }

      ev.clr();
      odd[br] = minim;
      br++;
      sum += minim;
      //cout << minim << endl;
      for (lli j = 0; j < odd.size(); ++j)
      {
        if (j != br)
          ev.pb(odd[j]);
      }
    }
  }
  cout << sum << endl;

  return 0;
}