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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> P;
typedef pair<ll, ll> Pll;
#define debug(var)  do{std::cout << #var << " : ";view(var);}while(0)
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

const int MOD = 1000000007;
const int INF = 1e9;

const int mod = 1000000007;
const int inf = 1e9;
#define PI acos(-1);
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int ddx[8] = {1,1,1,-1,-1,-1,0,0};
int ddy[8] = {0,1,-1,0,1,-1,1,-1};

int f[410][410] = {0};
int dp[410][410];
int n;
vector<int> a;
vector<int> sum;

ll rec(int i, int j) {
    if(f[i][j]==1) return dp[i][j];
    f[i][j] = 1;
    if(i == j) return 0;
    ll ret = 1e18;
    for(int k = i; k < j; k++) {
        chmin(ret, rec(i, k) + rec(k+1, j));
    }
    return dp[i][j] = ret + sum[j+1] - sum[i];
}

int main(){
    cin >> n;
    a.resize(n);
    for(int i = 0; i < n; i++) cin >> a[i];
    sum.assign(n+1,0);
    for(int i = 0; i < n; i++) sum[i+1] = sum[i] + a[i];
    ll ans = rec(0, n-1);
    cout << ans << endl;
}