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 Scala (2.11.7) (judge verdict: Wrong Answer)

import java.io.{InputStream, OutputStream, PrintStream}
import java.util.Scanner

object Main {

  def main(args: Array[String]): Unit = {
    val solver = new Solver(System.in, System.out)
    solver.solve()
  }

  class Solver(in: InputStream, out: OutputStream) {
    lazy val input = new Scanner(in)

    lazy val output = new PrintStream(out)

    def solve(): Unit = {
      val Array(n, m) = input.nextLine().split(" ").map(_.toInt)
      val ranges =
        1.to(m)
          .map { _ =>
            val Array(s, e) = input.nextLine().split(" ").map(_.toInt)
            (s, e)
          }

      var remainingRanges = ranges
      var count = 0
      while (remainingRanges.nonEmpty) {
        val (startCounts, endCounts) = (Array.fill(n)(0), Array.fill(n)(0))
        remainingRanges.foreach { case (s, e) =>
          startCounts(s - 1) += 1
          endCounts(e - 1) += 1
        }

        val values = Array.fill(n)(0)
        var overlaps = 0
        0.until(n).foreach { i =>
          overlaps += startCounts(i)
          overlaps -= endCounts(i)
          values(i) = overlaps
        }

        val max = values.max
        val i = values.indexOf(max)

        remainingRanges = remainingRanges.filterNot { case (s, e) =>
          s - 1 <= i && i < e - 1
        }

        count += 1
      }

      output.println(count)
    }

  }

}