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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.util.stream.Collectors;
import java.io.*;
import java.nio.charset.StandardCharsets;

public class Main {
	static class Tuple {
		long X, L;

		Tuple(long X, long L) {
			this.X = X;
			this.L = L;
		}

		long min() {
			return X - L;
		}

		long max() {
			return X + L;
		}
	}

	int range = 2000000000;
	List<Tuple> list;
	int N;
	int removeNum = 0;

	public static void main(String[] args) throws IOException {
		InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);
		BufferedReader in = new BufferedReader(reader);
		Main ins = new Main(in);
		ins.calc();
		ins.showResult();
	}

	Main(BufferedReader in) throws IOException {
		this.N = Integer.parseInt(in.readLine());
		this.list = new ArrayList<>(N);
		for (int i = 0; i < N; ++i) {
			String[] tokens = in.readLine().split(" ");
			list.add(new Tuple(Long.parseLong(tokens[0]), Long.parseLong(tokens[1])));
		}

	}

	boolean fillMap(Tuple tuple, boolean[] map) {
		boolean enable = true;
		int min = (int) Math.max(0, tuple.min());
		int max = (int) Math.min(range - 1, tuple.max());
		System.out.println(min + ", " + max);
		for (int i = min; i <= max; ++i) {
			if (map[i]) {
				enable = false;
				break;
			}
		}
		if (!enable) {
			return false;
		} else {
			Arrays.fill(map, min, max + 1, true);
			return true;
		}
	}

	void calc() {
		// まずLの大きさ順にソートする
		Collections.sort(list, new Comparator<Tuple>() {
			@Override
			public int compare(Tuple o1, Tuple o2) {
				return Long.compare(o1.max(), o2.max());
			}
		});

		long min = list.get(0).max();
		for(int i = 1; i < N; ++i) {
			Tuple tuple = list.get(i);
			if(min <= tuple.min()) {
				// 入れられる
				min = tuple.max();
			}else {
				// 入れられない
				++this.removeNum;
			}
		}
	}

	void showResult() {
		System.out.println((N-removeNum));
	}
}