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 __ <<" "<<
#define ___ <<" "
#define bash push_back
#define ALL(x) x.begin(),x.end()
#define int long long

struct IoSetup {
    IoSetup() {
        cin.tie(0);
        ios::sync_with_stdio(false);
        cout <<fixed<<setprecision(12);
        cerr <<fixed<<setprecision(12);
    }
}IoSetup;

typedef long long ll;
typedef pair < int , int >  pii;
static constexpr const signed int INF = 0x3f3f3f3f;
static constexpr const signed long long LINF = 0x3f3f3f3f3f3f3f3fLL;
static constexpr const signed int SMOD = 1000000007;
static constexpr const signed int NMOD = 998244353;
static constexpr const signed int dx[]={1,0,-1,0,1,1,-1,-1};
static constexpr const signed int dy[]={0,-1,0,1,-1,1,-1,1};

inline bool inside(int x,int y,int w,int h){return (x>=0 && y>=0 && x<w && y<h);}
template<class T>bool chmax(T &a, const T &b){return(a=max(a,b))==b;}
template<class T>bool chmin(T &a, const T &b){return(a=min(a,b))==b;}
inline int qp(int a,ll b,int mo){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}

int n;
int a[500];
int mem[500][500];
int sum[500];

int dp(int l, int r) {
    if(l==r) return 0;
    if(mem[l][r]) return mem[l][r];
    int ret = LINF;
    for(int i = l; i < r; i++) {
        chmin( ret, dp(l, i) + dp(i+1, r));
    }
    
    return mem[l][r] = ret + sum[r+1] - sum[l];
}


signed main() {

    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> a[i];
    }

    for(int i = 0; i < n; i++) {
        sum[i+1] = sum[i] + a[i];
    }
    
    cout << dp(0, n-1) << endl;
  
    return 0;
}