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: Time Limit Exceeded)

/**
 * @author derrick20
 */
import java.io.*;
import java.util.*;

public class Main {
    public static void main(String args[]) throws Exception {
        FastScanner sc = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int N = sc.nextInt();
        long W = sc.nextLong();
        long[] weight = new long[N];
        long[] value = new long[N];
        for (int i = 0; i < N; i++) {
            weight[i] = sc.nextLong();
            value[i] = sc.nextLong();
        }

        TreeMap<Long, Long> dp = new TreeMap<>(Collections.reverseOrder());
        dp.put(0L, 0L);
        long best = 0;
        // What's the best value achievable with weight W?
        // We must do FORWARD DP, due to the nature of treemaps containing keys
        for (int item = 0; item < N; item++) {
            long val = value[item];
            long wt = weight[item];

            TreeMap<Long, Long> tempMap = new TreeMap<>();
            for (long oldWt : dp.keySet()) {
                if (oldWt + wt > W) continue;
                // We can either add to the first state or something that was
                // reachable already
                // This statement makes me feel safer but it's actually NOT necessary,
                // since we're jsut maximizing everything

                long newVal = dp.get(oldWt) + val;
                long oldVal = 0;
                if (dp.containsKey(oldWt)) {
                // If there was an old value, maybe we won't update
                    oldVal = dp.get(oldWt);
                }
                if (newVal > oldVal) {
                    // Promptly update
                    tempMap.put(oldWt + wt, newVal);
                    best = Math.max(best, newVal);
                }
            }
            dp.putAll(tempMap);
        }
        out.println(best);
        out.close();
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        FastScanner(FileReader s) {
            br = new BufferedReader(s);
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        String nextLine() throws IOException { return br.readLine(); }

        double nextDouble() throws IOException { return Double.parseDouble(next()); }

        int nextInt() throws IOException { return Integer.parseInt(next()); }

        long nextLong() throws IOException { return Long.parseLong(next()); }
    }
}