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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_imports)]
use ascii::AsciiChar;
use ascii::AsciiStr;
use itertools::Itertools as _;
use permutohedron::LexicalPermutation as _;
use petgraph::algo::dijkstra;
use petgraph::algo::min_spanning_tree;
use petgraph::data::FromElements;
use petgraph::graph::NodeIndex;
use petgraph::graph::UnGraph;
use petgraph::unionfind::UnionFind;
use proconio::fastout;
use proconio::input;
use proconio::marker::Chars;
use proconio::marker::Usize1;
use std::cmp::max;
use std::cmp::min;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use superslice::Ext as _;

#[fastout]
fn main() {
    input! {
        N: usize,
        XLs: [(i64, i64); N],
    }

    // 間違い
    // let mut sorted_XLs: Vec<(i64, i64, i64)> = XLs.iter().map(|&(x, l)| (x - l, l, x)).collect();

    let mut sorted_XLs: Vec<(i64, i64, i64)> = XLs.iter().map(|&(x, l)| (x + l, l, x)).collect();

    sorted_XLs.sort();

    // 正当性が示せてないけど、いったん貪欲でやってみるか。
    // 一番左のものは必ず使うことにする
    // 手が左のものから順に見ていくが、横幅が小さいものを優先

    // 貪欲法なのは正しかったが、進め方が間違いだった。
    // 右手でどこまでいったかを見つつ左手がひっかからないように進めた場合、敷き詰めるような形になるから効率的ではないね。できるだけ右手が前に進まないようなものを選ぶべき。
    let mut ans: i64 = 0;
    let mut right = sorted_XLs[0].0;
    ans += 1;

    for i in 1..N {
        let (this_right, l, x) = sorted_XLs[i];
        let left = x - l;
        if right <= left {
            // 採用
            ans += 1;
            right = this_right;
        }
    }

    println!("{}", ans);
}