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 Scala (2.11.7) (judge verdict: Accepted)

import scala.io.Source
import scala.collection.mutable.Map

object Main {
  def main(args: Array[String]) = {
    val lines = Source.stdin.getLines.toStream
    val inSeq = lines.drop(1).head.split(' ').map(_.toLong)

    println(dp(inSeq))
  }

  def dp(input: IndexedSeq[Long]): Long = {
    val memo = Array.ofDim[(Long, Long)](input.length + 1, input.length + 1)

    (1 to input.length).foreach { n =>
      memo(n-1)(n) = input(n-1) -> 0
    }

    (2 to input.length).foreach { area =>
      var res = Long.MaxValue
      (2 to area).foreach { tgLen =>
        // cand
        var locMin = Long.MaxValue
        var cost = 0L
        ((area - tgLen + 1) to (area - 1)).reverse.foreach { split =>
          // println(area, tgLen, split)
          var cand = memo(area - tgLen)(split)._2 + memo(split)(area)._2
          if (cand <= locMin) {
            locMin = cand
            cost = memo(area - tgLen)(split)._1 + memo(split)(area)._1
          }
          // locMin = math.min(
          //   locMin, memo((area - tgLen, split))._2 + memo((split, area))._2
          // )
        }

        // println(area - tgLen, area, locMin)
        memo(area - tgLen)(area) = cost -> (locMin + cost)
      }
    }
    memo(0)(input.length)._2
  }
}