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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline

N = int(input())
A = list(map(int, input().split()))

dp = [[float('inf')]*N for _ in range(N)]
for i in range(N):
    dp[i][i] = 0


def dfs(l, r):
    """
    List[l:r]
    """
    #print("l,r->", l, r)
    # print(A[l:r])
    if dp[l][r-1] != float('inf'):
        return dp[l][r-1]
    if r-l == 1:
        # print("sole")
        return 0
    if r-l == 2:
        # print("double")
        dp[l][r-1] = A[l]+A[r-1]
        return dp[l][r-1]
    elif r-l <= 0:
        # print("err")
        return 0
    else:
        if l+2 >= r:
            #print("err->ret empty")
            return -1
        else:
            # print("update")
            dp[l][r-1] = min(dfs(l, v)+dfs(v, r)
                             for v in range(l+1, r))+sum(A[l:r])
            return dp[l][r-1]


dfs(0, N)
# print(dp)
#print("ans->", dp[0][-1])
print(dp[0][-1])