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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define inf 10000000000000
int dp[3001][3001];
int sum[3001][3001];

int solve(int i, int j) {
    if(i == j) return 0;
    if(dp[i][j] != -1) return dp[i][j];
    int least = inf;
    for(int k=i; k<j; k++) least = min(least, solve(i, k) + solve(k+1, j) + sum[i][j]);
    return dp[i][j] = least;
}

void prepro(vector<int>& vec, int N) {
    for(int i=1; i<=N; i++) {
        for(int j=i; j<=N; j++) sum[i][j] = vec[j] + ((i == j) ? 0 : sum[i][j-1]);
    }
}

int32_t main() {
    int N; cin>>N;
    vector<int> vec(N+1);
    for(int i=1; i<=N; i++) cin>>vec[i];
    memset(dp, -1, sizeof dp);
    prepro(vec, N);
    cout<<solve(1, N)<<endl;
    return 0;
}