The only difference between easy and hard versions is that you should complete all the projects in easy version but this is not necessary in hard version.

Polycarp is a very famous freelancer. His current rating is r units.

Some very rich customers asked him to complete some projects for their companies. To complete the i-th project, Polycarp needs to have at least a_i units of rating; after he completes this project, his rating will change by b_i (his rating will increase or decrease by b_i) (b_i can be positive or negative). Polycarp's rating should not fall below zero because then people won't trust such a low rated freelancer.

Is it possible to complete all the projects? Formally, write a program to check if such an order of the projects exists, that Polycarp has enough rating before starting each project, and he has non-negative rating after completing each project.

In other words, you have to check that there exists such an order of projects in which Polycarp will complete them, so he has enough rating before starting each project, and has non-negative rating after completing each project.

Input

The first line of the input contains two integers n and r (1 ≤ n ≤ 100, 1 ≤ r ≤ 30000) — the number of projects and the initial rating of Polycarp, respectively.

The next n lines contain projects, one per line. The i-th project is represented as a pair of integers a_i and b_i (1 ≤ a_i ≤ 30000, -300 ≤ b_i ≤ 300) — the rating required to complete the i-th project and the rating change after the project completion.

Output

Print "YES" or "NO".

Examples

Input


3 4
4 6
10 -2
8 -1


Output


YES


Input


3 5
4 -5
4 -2
1 3


Output


YES


Input


4 4
5 2
5 -3
2 1
4 -2


Output


YES


Input


3 10
10 0
10 -10
30 0


Output


NO

Note

In the first example, the possible order is: 1, 2, 3.

In the second example, the possible order is: 2, 3, 1.

In the third example, the possible order is: 3, 1, 4, 2.
// Don't place your source in a package
import javax.swing.*;
import java.lang.reflect.Array;
import java.text.DecimalFormat;
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
import java.util.stream.Stream;


// Please name your class Main
public class Main {
    static FastScanner fs=new FastScanner();
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next() {
            while (!st.hasMoreElements())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        int Int() {
            return Integer.parseInt(next());
        }

        long Long() {
            return Long.parseLong(next());
        }

        String Str(){
            return next();
        }
    }


    public static void main (String[] args) throws java.lang.Exception {
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

        int T=1;
        for(int t=0;t<T;t++){
            int n=Int();
            int r=Int();
            int A[][]=new int[n][2];
            for(int i=0;i<n;i++){
                A[i][0]=Int();
                A[i][1]=Int();
            }

            Solution sol=new Solution(out);
            sol.solution(A,r);
        }
        out.close();

    }


    public static int Int(){
        return fs.Int();
    }
    public static long Long(){
        return fs.Long();
    }
    public static String Str(){
        return fs.Str();
    }

}






class Solution{
    PrintWriter out;
    public Solution(PrintWriter out){
        this.out=out;
    }





    public void solution(int A[][],int sum){
        Arrays.sort(A,(a,b)->{
            return a[0]-b[0];
        });
        PriorityQueue<int[]>pq=new PriorityQueue<>((a,b)->{
            return b[1]-a[1];
        });

        for(int i=0;i<A.length;i++){
            if(A[i][0]<=sum){
                pq.add(A[i]);
            }
            else{
                out.println("NO");
                return;
            }
            while(pq.size()>0&&pq.peek()[1]>=0)sum+=pq.poll()[1];
        }

        //System.out.println(sum);

        PriorityQueue<int[]>pq1=new PriorityQueue<>((a,b)->{
            int c1=-a[1];
            int c2=-b[1];
            if(a[0]==b[0])return c1-c2;

            int dif=Math.abs(a[0]-b[0]);
            if(a[0]>b[0]){
                return (c1-dif)-c2;
            }
            else{
                return c1-(c2-dif);
            }
        });


        while(pq.size()>0)pq1.add(pq.poll());
        while(pq1.size()>0){
            int top[]=pq1.poll();
            //System.out.println(Arrays.toString(top));
            if(sum<top[0]){
                out.println("NO");
                return;
            }
            sum+=top[1];
        }
        if(sum<0){
            out.println("NO");
            return;
        }
        out.println("YES");

    }






}







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                      free bug dog
*/


