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)

/*
      @author: Charan Sai
*/
#include <bits/stdc++.h>
using namespace std;

#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define int long long
#define ppi pair<int, int>
#define endl "\n"
#define nl cout << "\n"
#define deb(x) cout << #x << " " << x << endl;
#define itr(i, n) for(int i = 0; i < n; ++i)
#define itr1(i, a, n) for(int i = a; i < n; ++i)

const int INF = 1e10;
const int MAX_N = 410;

int dp[MAX_N][MAX_N];
int sum[MAX_N][MAX_N];

int top_down(int i, int j) {
    if(i == j) return 0;
    if(i > j) return INT_MAX;

    if(dp[i][j] != -1) return dp[i][j];
    int res = INF;
    for(int k = i; k <= j-1; ++k) {
        res = min(res, top_down(i, k) + top_down(k+1, j) + sum[i][j]);
    }
    return (dp[i][j] = res);
}

int32_t main() {
    memset(dp, -1, sizeof(dp));
    memset(sum, 0, sizeof(sum));
    IOS;
    int n; cin >> n;
    vector<int> a(n);
    for(int &x : a) cin >> x;
    for(int i = 0; i < n; ++i) {
        for(int j = i; j < n; ++j) {
            if(j == 0) sum[i][j] = a[0];
            sum[i][j] = a[j] + sum[i][j-1];
        }
    }
    cout << top_down(0, n-1) << endl;
}