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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

func nextInt(sc *bufio.Scanner) int {
	sc.Scan()
	t, _ := strconv.Atoi(sc.Text())
	return t
}

type point struct {
	x, l int
}

type ps []point

func (p ps) Len() int {
	return len(p)
}
func (p ps) Swap(i, j int) {
	p[i], p[j] = p[j], p[i]
}
func (p ps) Less(i, j int) bool {
	return p[i].x < p[j].x
}

func dup(a, b point) bool {
	//fmt.Println("dup en: a", a, "b", b)
	A, B := float64(a.x-a.l)+0.0001, float64(a.x+a.l)-0.0001
	X, Y := float64(b.x-b.l)+0.0001, float64(b.x+b.l)-0.0001
	return X <= B && A <= Y
}

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	N := nextInt(sc)
	// X := make([]int, N)
	// L := make([]int, N)
	points := make([]point, N)
	for i := 0; i < N; i++ {
		points[i].x = nextInt(sc)
		points[i].l = nextInt(sc)
	}
	sort.Sort(ps(points))
	// fmt.Println("N", N, "points", points)
	for i := 0; i+1 < N; i++ {
		if dup(points[i], points[i+1]) {
			// fmt.Println("d00")
			// ロボットiとロボットi+1は重なる
			// どっちを取り除いたほうが良いか判定
			if i+2 < N {
				// fmt.Println("d01")
				if dup(points[i], points[i+2]) {
					// fmt.Println("d02")
					// iはi+2とも重なる
					// iを除く
					points[i].x = -1
				} else {
					// fmt.Println("d03")
					// iはi+2とは重ならない
					// i+1を除く
					points[i+1] = points[i]
					points[i].x = -1
					points[i].l = -1
				}
			} else {
				// fmt.Println("d04")
				points[i+1].x = -1
				points[i+1].l = -1
			}
		}
	}
	ans := 0
	// fmt.Println(points)
	for i := 0; i < len(points); i++ {
		if points[i].x != -1 {
			ans++
		}
	}
	fmt.Println(ans)
}