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 10^5

1 \leq w_i \leq W

1 \leq v_i \leq 10^9

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

5 5
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000

Sample Output 2

5000000000

The answer may not fit into a 32-bit integer type.

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: Wrong Answer)

//package src;
import java.io.*;
//import javafx.util.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
	public static void main (String[] args)  {
//		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//		int t = Integer.parseInt(br.readLine());
//		while(t-->0)
//		{
//		  int totItems = Integer.parseInt(br.readLine()); //no.of items.
//		  int n = Integer.parseInt(br.readLine()); //capacity of knapsack.
//		  int val[] = new int[totItems]; //values of items.
//		  int wt[] = new int[totItems]; //weights of items.
//		  
//		  
//		  for(int i=0;i<totItems;i++){
//            String s1[]= br.readLine().split(" ");
//		    val[i] = Integer.parseInt(s1[i]);
//            wt[i] = Integer.parseInt(s1[i]);
//          }
		    
		   // String s2[]= br.readLine().split(" ");
		//  for(int i=0;i<;i++)
		  //  wt[i] = Integer.parseInt(s2[i]);
		Scanner sc =new Scanner(System.in);
		int totItems=sc.nextInt();
		int n=sc.nextInt();
		int val[]=new int [totItems];
		int wt[]=new int [totItems];
		for(int i=0;i<totItems;i++) {
		 wt[i]=sc.nextInt();
		val[i]=sc.nextInt();
		}
		    findMaxProfit(totItems,n,val,wt);
		  
		
	}
	
	static void findMaxProfit(int totItems,int n,int val[],int wt[])
	{
	    int dp[][] = new int[n+1][totItems+1];
	   //IMP:-Top X-axis - > Capacity from 0....to n.    Draw graph to visualise.
	    //    Y-axis - > Items from 0...to totItems.
	    
	    //First row and first column is filled with 0's.
	    
	    for(int i=1;i<=n;i++)
	    {
	        for(int j=1;j<=totItems;j++)
	        {
	           //If the weight of the item is less than or equal to the capacity.
	            if(wt[j-1]<=i) {
	   /*IMP:-  Take the max of(val of cur_item + (DP answer -> [cur_capacity 
	            -weight of cur_item][above item/cell]) 
	                                AND 
	            (DP answer -> dp[same capacity][above item/cell] */
 
	            dp[i][j] = Math.max(val[j-1]+dp[i-wt[j-1]][j-1] , dp[i][j-1]);   }
	            
	            else
	            dp[i][j] = dp[i][j-1];
	        }
	    }
	    //Gives the max value obtained for the capacity n considering all the items.
	    System.out.println(dp[n][totItems]) ;  
	}
}