Problem:
Score : 100 points

Problem Statement

There are N items, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i.

Taro has decided to choose some of the N items and carry them home in a knapsack.
The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W.

Find the maximum possible sum of the values of items that Taro takes home.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq W \leq 10^9

1 \leq w_i \leq W

1 \leq v_i \leq 10^3

Input

Input is given from Standard Input in the following format:

N W
w_1 v_1
w_2 v_2
:
w_N v_N

Output

Print the maximum possible sum of the values of items that Taro takes home.

Sample Input 1

3 8
3 30
4 50
5 60

Sample Output 1

90

Items 1 and 3 should be taken.
Then, the sum of the weights is 3 + 5 = 8, and the sum of the values is 30 + 60 = 90.

Sample Input 2

1 1000000000
1000000000 10

Sample Output 2

10

Sample Input 3

6 15
6 5
5 6
6 4
6 6
3 5
7 2

Sample Output 3

17

Items 2, 4 and 5 should be taken.
Then, the sum of the weights is 5 + 6 + 3 = 14, and the sum of the values is 6 + 6 + 5 = 17.

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

require 'pp'

# 空白区切の入力値を数値の配列で返却する
def gets_i_list()
  gets.chomp.split(" ").map(&:to_i)
end

N, W = gets_i_list
wv_list = []
N.times do
  wv_list.push gets_i_list
end

# 最大容量以下で、容量ごとの価値の最大値を格納する
dp = Array.new(W + 1, 0)

wv_list.each do |w, v|
  # 許容できる容量(wからW)の範囲の価値を更新する
  # uptoを使うと、ループで更新した値を参照してしまうため、downtoを使う
  W.downto(w) do |i|
    # 現在の値(dp[i])より、　itemを追加した場合の価値(dp[i - w] + v)が大きければ更新する
    if dp[i] < dp[i - w] + v
      dp[i] = dp[i - w] + v 
    end
  end
end
puts dp.max