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>

#define int long long int
#define mp make_pair
using namespace std;

int n;
vector <int> v(402);
vector <int> pre(402);
int dp[402][402];

int fun(int start , int end)
{
    if(start == end){
        return 0;
    }
    if(start + 1 == end){
        return v[start] + v[end];
    }
    if(dp[start][end] != 1e18){
        return dp[start][end];
    }
    for (int i = start; i < end; ++i) {
        dp[start][end] = min(dp[start][end] , fun(start , i) + fun(i+1 , end)) + pre[end] - pre[start - 1] ;
    }
    return dp[start][end];
}

int32_t main() {
    scanf("%lld", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%lld", &v[i]);
        pre[i] = pre[i-1] + v[i];
    }
    for (int i = 0; i < 402; ++i) {
        for (int j = 0; j < 402; ++j) {
            dp[i][j] = 1e18;
        }
    }
    int ans = fun(1 , n);
    printf("%lld",ans);
    return 0;
}