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 Ruby (2.3.3) (judge verdict: Wrong Answer)

require 'pp'
N = ARGF.gets.to_i
A = ARGF.gets.split(" ").map(&:to_i)

dp = Array.new(N) do Array.new(N, 1000) end
weight = Array.new(N) do Array.new(N, 0) end

def pri(weight)
  puts weight.map{|r| r.map{|x| "%04d"%x}.join(",")}.join("\n")
end

for w in 1..N
  for l in 0..N-w
    r = l+w-1
    if w == 1
      dp[l][r] = 0
      weight[l][r] = A[l]
      next
    end

    if w == 2
      dp[l][r] = A[l] + A[r] # dp[l][l+1]
      weight[l][r] = A[l] + A[r]
      next
    end

    weight[l][r] = weight[l+1][r] + A[l]

    for wl in 1...w
      n = dp[l][l+wl-1] + dp[l+wl][r] + weight[l][l+wl-1] + weight[l+wl][r]

      if dp[l][r] > n
        dp[l][r] = n
      end
    end
  end
end

puts dp[0][N-1]