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;

int A[404];
long long dp[404][404], S[404][404];

long long solve(int left, int right)
{
    if(left == right)
        return 0;
    
    if(dp[left][right] != -1)
        return dp[left][right];
    
    long long ans = (long long) 1e18;
    
    for(int i = left+1; i <= right; i++)
        ans = min(ans, solve(left, i-1) + solve(i, right) + S[left][i-1] + S[i][right]);
    
    return dp[left][right] = ans;
}

int main() {

    ios::sync_with_stdio(0);
    cin.tie(NULL), cout.tie(NULL);
    
    int N;
    cin >> N;
    
    for(int i=1; i<=N; i++)
        cin >> A[i];
    
    for(int i=1; i<=N; i++)
        for(int j=i; j<=N; j++)
            S[i][j] = S[i][j-1] + A[j];
    
    memset(dp, -1, sizeof(dp));
    
    cout << solve(1, N) << "\n";
    return 0;
}