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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Scanner;

public class Main {

    static int N = 0;
    static int W = 0;
    static int[] wi;
    static int[] vi;
    static long memo[][]; //関数の結果を保存するための配列

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        N = Integer.parseInt(sc.next());
        W = Integer.parseInt(sc.next());
        wi = new int[N];
        vi = new int[N];
        memo = new long[N+1][W+1];

        for (int i = 0; i < N; i++) {

            wi[i] = Integer.parseInt(sc.next());
            vi[i] = Integer.parseInt(sc.next());
        }
        sc.close();

        System.out.println(knap(0,W));

    }

    //全探索する再帰
    /**
     * @param i i番目の商品
     * @param j 重さ
     * @return 最大値をとりうる場合の価値の総和、i番目の商品が重さjのときに使われたらその場所にメモされる
     */
    public static long knap(int i,int j) {

        if (memo[i][j] != 0) {

            return memo[i][j];
        }

        long res = 0;

        if (i == N) { //商品がもうないので0を代入する

            res = 0;
        } else if (j < wi[i]) { //容量不足なので入れない処理だけをする

            res = knap(i+1, j);
        } else { //入れると最大になるとは限らないので入れる時と入れない時の大きい方を取る

            res = Math.max(knap(i+1, j), knap(i+1, j-wi[i]) + vi[i]);
        }
      
        return memo[i][j] = res;
    }

}