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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

# https://kyopro-friends.hatenablog.com/entry/2019/01/12/231035


def div_cost(left, right):
    # 区間[left, right]が一匹にまとまっているとき、それを分解するために必要な最小コスト
    if bl_calculated[left][right]:
        return memo[left][right]
    bl_calculated[left][right] = True
    if left == right:
        memo[left][right] = 0
        return memo[left][right]
    divided_to_alone_cost = INF
    # divided_to_alone_cost: 区間[left, right]を分解してできた2つの区間それぞれをaの各要素まで分解し切るのに必要なコスト
    for mid in range(left, right):
        divided_to_alone_cost = min(divided_to_alone_cost, div_cost(left, mid) + div_cost(mid + 1, right))
    memo[left][right] = divided_to_alone_cost + (c_sum[right] - c_sum[left - 1])
    # (c_sum[right] - c_sum[left - 1]): 区間[left, right]を分解するのにかかるコスト
    return memo[left][right]


INF = float('inf')

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

c_sum = [0]
for aa in a:
    c_sum.append(c_sum[-1] + aa)
# 累積和c_sum

memo = [[0 for _ in range(N + 1)] for _ in range(N + 1)]
# memo[left][right] = (区間[left, right]に相当するスライムが一匹にまとまっているとき、それを分解するために必要な最小コスト)

bl_calculated = [[False for _ in range(N + 1)] for _ in range(N + 1)]
# memo[left][right]に最小コストが代入済かどうかのBoolean

print(div_cost(1, N))