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 <cstdio>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;

typedef long long int64;
constexpr int p = 1e9+7;
constexpr long long M = 0x777777777777777777777777777777777777777LL;

int N;
int a[408];
int64 dp[408][408];

int64 solve(int begin, int end) {
    if(begin==end) {
        return 0;
    }
    if(!dp[begin][end]) {
        int64 ret = M;
        for(int i=begin; i<end; ++i) {
            ret = min(ret, solve(begin, i) + solve(i+1, end));
        }
        for(int i=begin; i<=end; ++i) {
            ret += a[i];
        }
        dp[begin][end] = ret ^ M;
    }
    return dp[begin][end] ^ M;
}

int main() {
	scanf("%d", &N);
    for(int n=0; n<N; ++n) {
        scanf("%d", &a[n]);
    }
    printf("%lld\n", solve(0, N-1));
	return 0;
}