Problem:
Score : 200 points

Problem Statement

In a factory, there are N robots placed on a number line.
Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative.

We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect.
Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints.

Find the maximum number of robots that we can keep.

Constraints

1 \leq N \leq 100,000

0 \leq X_i \leq 10^9 (1 \leq i \leq N)

1 \leq L_i \leq 10^9 (1 \leq i \leq N)

If i \neq j, X_i \neq X_j.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 L_1
X_2 L_2
\vdots
X_N L_N

Output

Print the maximum number of robots that we can keep.

Sample Input 1

4
2 4
4 3
9 3
100 5

Sample Output 1

3

By removing Robot 2, we can keep the other three robots.

Sample Input 2

2
8 20
1 10

Sample Output 2

1

Sample Input 3

5
10 1
2 1
4 1
6 1
8 1

Sample Output 3

5

Solution in Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.*;
import java.math.*;
import java.util.*;
public class Main { 
 
	public static void main(String[] args) { 
		FastScanner I = new FastScanner(); //Input
		OutPut O = new OutPut(); //Output
		long ans = 0;
		int N = I.nextInt();
		long[][] rob = new long[N][2]; //[Position,Wing-span]
		boolean[] vis = new boolean[N+1];
		long up = 0; //Current largest X coordinate occupied
		for (int i = 0; i < N; i++) {
			rob[i][0] = I.nextLong();
			rob[i][1] = I.nextLong()-1; //End-points excluded
		}
		Arrays.sort(rob,(a,b)->Long.compare(a[0],b[0]));
		up = rob[0][0]+rob[0][1];
		vis[0]=true;
		int cur = 0; //Which robot was most recently included in optimal set
		for (int i = 1; i < N; i++) {
			long curlo = rob[i][0]-rob[i][1];
			long curhi = rob[i][0]+rob[i][1];
			if (curlo<=up) { //if overlap
				if (curhi<up) {
					up=curhi; //Including current, excluding previous
					vis[cur]=false;
					vis[i]=true;
					cur=i;
				} 
			}else {
				vis[i]=true;
				up=curhi;
			}
		}
		for (int i = 0; i < N; i++) if (vis[i]) ans++;
		O.pln(ans);
	}
	public static long ceil(long num, long den) {long ans = num/den; if (num%den!=0) 
	ans++; return ans;}
	public static long GCD(long a, long b) {
		if (a==0||b==0) return Math.max(a,b);
		return GCD(Math.min(a, b),Math.max(a, b)%Math.min(a, b));
	}
	public static long FastExp(long base, long exp, long mod) {
		long ans=1;
		while (exp>0) {
			if (exp%2==1) ans*=base;
			exp/=2;
			base*=base;
			base%=mod;
			ans%=mod;
		}
		return ans;
	}
	public static long ModInv(long num,long mod) {return FastExp(num,mod-2,mod);}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());};
	}
	static class OutPut{
		PrintWriter w = new PrintWriter(System.out);
		void pln(int x) {w.println(x);w.flush();}
		void pln(long x) {w.println(x);w.flush();}
		void pln(String x) {w.println(x);w.flush();}
		void pln(char x) {w.println(x);w.flush();}
		void pln(StringBuilder x) {w.println(x);w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
		void p(StringBuilder x) {w.print(x);w.flush();}
	}
}