Problem:
Knapsack Problem

You have N kinds of items that you want to put them into a knapsack. Item i has value vi and weight wi.

You want to find a subset of items to put such that:

The total value of the items is as large as possible.

The items have combined weight at most W, that is capacity of the knapsack.

You can select as many items as possible into a knapsack for each kind.

Find the maximum total value of items in the knapsack.

Input

N W
v1 w1
v2 w2
:
vN wN

The first line consists of the integers N and W. In the following lines, the value and weight of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Constraints

1 ≤ N ≤ 100

1 ≤ vi ≤ 1000

1 ≤ wi ≤ 1000

1 ≤ W ≤ 10000

Sample Input 1

4 8
4 2
5 2
2 1
8 3

Sample Output 1

21

Sample Input 2

2 20
5 9
4 10

Sample Output 2

10

Sample Input 3

3 9
2 1
3 1
5 2

Sample Output 3

27

Solution in Scala (judge verdict: Wrong Answer)

object Main {
  def main(args: Array[String]): Unit = {
    val (num, limit) = {
      val nw = io.StdIn.readLine().split(" ").map(_.toInt)
      (nw(0), nw(1))
    }
    val items = new Array[(Int, Int)](num)
    for (i <- 0 until num) {
      val vw = io.StdIn.readLine().split(" ").map(_.toInt)
      items(i) = (vw(0), vw(1))
    }

    val dp = solve(limit, items)
    println(dp(num)(limit))
  }

  def solve(limit: Int, items: Array[(Int, Int)]): Array[Array[Int]] = {
    val dp = Array.ofDim[Int](items.length + 1, limit + 1)

    items.zipWithIndex.foreach { case ((v, w), i) =>
      for (l <- 1 to limit) {
        if (w <= l && dp(i)(l) < dp(i)(l - w) + v) {
          dp(i + 1)(l) = dp(i)(l - w) + v
        } else {
          dp(i + 1)(l) = dp(i)(l)
        }
      }
    }
    dp
  }
}