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.

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import java.util.Collections;
import java.util.ArrayList;
import java.util.Scanner;

public class Main
{
	public static void main(String[] args)
	{
		//First we read all of our points. Each can be treated as an open interval centered
		//at the given location, with radius equal to the weight, and our goal is to find
		//the largest possible set of disjoint intervals
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		ArrayList<Pair> points = new ArrayList<Pair>();
		
		int maxLoc = -1;
		for(int i = 0; i < n; i++)
		{
			int loc = sc.nextInt();
			int weight = sc.nextInt();
			points.add(new Pair(loc, weight));
			if(loc > maxLoc)
				maxLoc = loc;
		}
		sc.close();
		
		//Next we sort our intervals from left to right by their right endpoints (location + weight)
		Collections.sort(points);
		
		//We want to do dynamic programming on an array. But we can't make an array of size 10^9.
		//So we store a compressed version of our array. We only need to store the values at the
		//right endpoints of intervals.
		int[] cliqueSizes = new int[n];
		for(int i = 0; i < n; i++)
		{
			//System.out.println("Point: " + Integer.toString(points.get(i).elem1) + ", " + Integer.toString(points.get(i).elem2));
			int left = points.get(i).left();
			
			int newSize = 1;
			if(left >= points.get(0).elem1 + points.get(0).elem2) //Check if we can include previous points
			{
				int prevIndex = prevSize(points, left, 0, n - 1);
				//System.out.println("Prev index: " + Integer.toString(prevIndex));
				newSize += cliqueSizes[prevIndex];
			}
			
			//Either choose the current point and use the best set for everything to the left of it...
			if(i == 0 || newSize > cliqueSizes[i - 1])
				cliqueSizes[i] = newSize;
			//Or decide not to use the current point.
			else
				cliqueSizes[i] = cliqueSizes[i - 1];
			
			//System.out.println("Clique size: " + Integer.toString(cliqueSizes[i]) + "\n");
		}
		
		System.out.println(cliqueSizes[n - 1]);
	}
	
	//To look up a previously computed value, we have to do a binary search on our list of points
	//to get the largest index that is smaller than or equal to the given value. Then we look up that index
	//in cliqueSizes.
	public static int prevSize(ArrayList<Pair> points, int left, int start, int end)
	{
		//Invariant: Right endpoint of point at start index is smaller than the given value,
		//           Right endpoint of point at end index is at least equal to the given value
		//System.out.println(Integer.toString(start) + ", " + Integer.toString(end));
		if(start == end)
			return start;
		else if(start + 1 == end)
		{
			if(points.get(end).right() <= left)
				return end;
			else
				return start;
		}
		else
		{
			int currLoc = (start + end)/2;
			if(points.get(currLoc).right() <= left)
				return prevSize(points, left, currLoc, end);
			else
				return prevSize(points, left, start, currLoc);
		}
	}
	
	static class Pair implements Comparable<Pair>
	{
		int elem1;
		int elem2;
		
		public Pair(int e1, int e2)
		{
			elem1 = e1;
			elem2 = e2;
		}
		
		public int compareTo(Pair other)
		{
			return ((Integer)(elem1 + elem2)).compareTo(other.elem1 + other.elem2);
		}
		
		public int left()
		{
			return elem1 - elem2;
		}
		
		public int right()
		{
			return elem1 + elem2;
		}
	}
}
