The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph G. It is required to find a subset of vertices C of the maximum size such that any two of them are connected by an edge in graph G. Sounds simple, doesn't it? Nobody yet knows an algorithm that finds a solution to this problem in polynomial time of the size of the graph. However, as with many other NP-complete problems, the clique problem is easier if you consider a specific type of a graph.

Consider n distinct points on a line. Let the i-th point have the coordinate xi and weight wi. Let's form graph G, whose vertices are these points and edges connect exactly the pairs of points (i, j), such that the distance between them is not less than the sum of their weights, or more formally: |xi - xj| ≥ wi + wj.

Find the size of the maximum clique in such graph.

Input

The first line contains the integer n (1 ≤ n ≤ 200 000) — the number of points.

Each of the next n lines contains two numbers xi, wi (0 ≤ xi ≤ 109, 1 ≤ wi ≤ 109) — the coordinate and the weight of a point. All xi are different.

Output

Print a single number — the number of vertexes in the maximum clique of the given graph.

Examples

Input

4
2 3
3 1
6 1
0 2


Output

3

Note

If you happen to know how to solve this problem without using the specific properties of the graph formulated in the problem statement, then you are able to get a prize of one million dollars!

The picture for the sample test.

<image>
/*
 *
 * @author Mukesh Singh
 *
 */

import java.io.*;
import java.util.*;
import java.text.DecimalFormat;
@SuppressWarnings("unchecked")
public class AB {	
	
	//solve test cases
	void solve() throws Exception {
		int n = in.nextInt();
		Node[] ar = new Node[n];
		for(int i = 0 ; i < n ; ++i ) ar[i] = new Node(in.nextInt(),in.nextInt());
		Arrays.sort(ar);
		int last = 0 ; int res = 1 ; 
		for(int i = 1 ; i < n; ++i ){
			//System.out.println(ar[i].sum);
			if(ar[i].x-ar[i].w >= ar[last].x+ar[last].w){
				last = i ; ++res;
			}
		}
		System.out.println(res);
	}
	class Node implements Comparable<Node>{
		int x ; int w ; int sum ; Node(int a , int b){x = a ; w = b ; sum= a+b;}
		public int compareTo(Node n1){
			return sum-n1.sum;
		}
	}
	//@ main function
	public static void main(String[] args) throws Exception {
		new AB();
	}
	
	InputReader in;
	PrintStream out ;
	DecimalFormat df ;
	AB() {
		try {
			File defaultInput = new File("file.in");
			if (defaultInput.exists()) 
				in = new InputReader("file.in");
			else 
				in = new InputReader();
			defaultInput = new File("file.out");
			if (defaultInput.exists()) 
				out = new PrintStream(new FileOutputStream("file.out"));
			else
				out = new PrintStream(System.out);
			df = new DecimalFormat("######0.00");
			solve();
			out.close();
		} 
		catch (Exception e) {
			e.printStackTrace();
			System.exit(261);
		}
	}
	
	class InputReader {
		BufferedReader reader;
		StringTokenizer tokenizer;
		
		InputReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
		}
		
		InputReader(String fileName) throws FileNotFoundException {
			reader = new BufferedReader(new FileReader(new File(fileName)));
		}
		
		String readLine() throws IOException {
			return reader.readLine();
		}
		
		String nextToken() throws IOException {
			while (tokenizer == null || !tokenizer.hasMoreTokens())
				tokenizer = new StringTokenizer(readLine());
			return tokenizer.nextToken();
		}
		
		boolean hasMoreTokens() throws IOException {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				String s = readLine();
				if (s == null)
					return false;
				tokenizer = new StringTokenizer(s);
			}
			return true;
		}
		
		int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(nextToken());
		}
		
		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(nextToken());
		}
		
		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(nextToken());
		}
	}
}
