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>
#define rep(i, n) for (lli i = 0; i < (n); i++)
#define rrep(i, n) for (lli i = (n)-1; i >= 0; i--)
using namespace std;
using lli = long long int;
#define int long long int
void YESNO(bool), YesNo(bool);
template <class T1, class T2>
bool chmin(T1 &l, const T2 &r);
template <class T1, class T2>
bool chmax(T1 &l, const T2 &r);
constexpr int M = 405;
lli dp[M][M] = {};
lli sum[M][M] = {};
void solve(long long N, std::vector<long long> a)
{
  rep(i, M) rep(j, M) dp[i][j] = 1e18;
  rep(i, N) dp[i][i + 1] = 0;
  rep(i, N) for (int j = i + 1; j <= N; j++)
  {
    // sum of a[j, i)
    sum[i][j] = sum[i][j - 1] + a[j - 1];
  }
  for (int r = 2; r <= N; r++)
  {
    rep(i, N)
    {
      // [i, i+r)
      for (int k = i + 1; k < i + r && i + r <= N; k++)
        chmin(dp[i][i + r], dp[i][k] + dp[k][i + r] + sum[i][k] + sum[k][i + r]);
    }
  }
  cout << dp[0][N] << endl;
}

signed main()
{
  long long N;
  scanf("%lld", &N);
  std::vector<long long> a(N);
  for (int i = 0; i < N; i++)
  {
    scanf("%lld", &a[i]);
  }
  solve(N, std::move(a));
  return 0;
}

// -- lib
void YESNO(bool b) { cout << (b ? "YES" : "NO") << endl; }
void YesNo(bool b) { cout << (b ? "Yes" : "No") << endl; }

template <class T1, class T2>
bool chmin(T1 &l, const T2 &r)
{
  return (l > r) ? (l = r, true) : false;
}

template <class T1, class T2>
bool chmax(T1 &l, const T2 &r)
{
  return (l < r) ? (l = r, true) : false;
}