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 C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Collections.Generic;
using System.Threading.Tasks;

class Program
{
	//Main
	static void Main(string[] args)
	{
		//読み込む
		//ロボット（区間）の総数（配列の要素数）
		int N = int.Parse(Console.ReadLine());
		//腕の終点（X+L）をkeyとし、腕の長さ(L)をValueとするディクショナリ
		SortedDictionary<long, long> roboDic = new SortedDictionary<long, long>();

		//Xi, Liを読み込む
		for(int i=0; i<N; i++){
				//読み込み
				string[] line = Console.ReadLine().Split(' ');
				long Xi = long.Parse(line[0]);
				long Li = long.Parse(line[1]);

				//ディクショナリに格納
				roboDic.Add((Xi+Li), Li);
		}

		//終点の順に、始点と終点とを配列に格納
		int j = 0;
		//始点を格納
		long[] starts = new long[N];
		//終点を格納
		long[] ends = new long[N];

		foreach(KeyValuePair<long,long> pair in roboDic){
				long L = pair.Value;
				long X = pair.Key - L;
				starts[j] = X-L;
				ends[j] = X+L;
				j++;
		}

		/*
		左端のロボットを残す最適解が存在するので、貪欲法で解けるとのこと
		*/
		//現在の終点よりも右にあれば、加える
		long curEnd = ends[0];
		int cnt = 1;
		for(int i=0; i<N; i++){
				//i番目のロボの始点が被らない場所に有る場合
				if(starts[i] >= curEnd){
						//カウント
						cnt ++;
						//現在の終点を更新
						curEnd = ends[i];
				}
		}

		//出力
		Console.WriteLine(cnt);
	}
}