Problem:
Score : 400 points

Problem Statement

There are N islands lining up from west to east, connected by N-1 bridges.

The i-th bridge connects the i-th island from the west and the (i+1)-th island from the west.

One day, disputes took place between some islands, and there were M requests from the inhabitants of the islands:

Request i: A dispute took place between the a_i-th island from the west and the b_i-th island from the west. Please make traveling between these islands with bridges impossible.

You decided to remove some bridges to meet all these M requests.

Find the minimum number of bridges that must be removed.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq a_i < b_i \leq N

All pairs (a_i, b_i) are distinct.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the minimum number of bridges that must be removed.

Sample Input 1

5 2
1 4
2 5

Sample Output 1

1

The requests can be met by removing the bridge connecting the second and third islands from the west.

Sample Input 2

9 5
1 8
2 7
3 5
4 6
7 9

Sample Output 2

2

Sample Input 3

5 10
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5

Sample Output 3

4

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use std::collections::BTreeSet;

fn main() {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.trim().split(" ");
    let n: usize = iter.next().unwrap().parse().unwrap();
    let m: usize = iter.next().unwrap().parse().unwrap();
    let mut v: Vec<(u32, u32)> = Vec::with_capacity(m);
    for _ in 0..m {
        let mut buf = String::new();
        io::stdin().read_line(&mut buf).unwrap();
        let mut iter = buf.trim().split(" ");
        let a: u32 = iter.next().unwrap().parse().unwrap();
        let b: u32 = iter.next().unwrap().parse().unwrap();
        v.push((a, b));
    }
    v.sort_by_key(|&x| x.1);
    let mut s: BTreeSet<u32> = BTreeSet::new();
    for (a, b) in v {
        if s.iter().any(|&x| { a <= x && x < b }) {
            continue;
        }
        s.insert(b - 1);
    }
    println!("{}", s.len());
}