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 <cstdio>
#include <cstdlib>
#include <cassert>
#include <algorithm>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define repi(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define repdi(i,a,b) for(ll i=(a)-1;i>=(b);--i)
#define repd(i,a) repdi(i,a,0)
#define itr(it,a) for( auto it = (a).begin(); it != (a).end(); ++it )
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()

using ll = long long;
using P = std::pair<ll, ll>;

constexpr ll INF = 1ll<<60;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<class S, class T>
std::ostream& operator<< ( std::ostream& out, const std::pair<S,T>& a )
{ std::cout << '(' << a.first << ", " << a.second << ')'; return out; }

template<class T>
std::ostream &operator<< ( std::ostream& out, const std::vector<T>& a )
{ std::cout << '['; rep( i, a.size() ){ std::cout << a[i]; if( i != a.size()-1 ) std::cout << ", "; } std::cout << ']'; return out; }

ll N;
ll a[410];
ll dp[410][410];
ll S[410];

int main()
{
  std::cin >> N;

  rep( i, N )
  {
    std::cin >> a[i];

    S[i+1] = S[i]+a[i];
  }

  rep( i, N+1 ) rep( j, N+1 )
    dp[i][j] = INF;

  rep( i, N )
    dp[i][i] = 0;

  repi( l, 1, N+1 ) rep( i, N-l ) repi( k, i, i+l )
    chmin( dp[i][i+l], dp[i][k]+dp[k+1][i+l]+S[i+l+1]-S[i] );

  std::cout << dp[0][N-1] << std::endl;

  return 0;
}