Problem:
Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands.
For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands.
Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N.
However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

3 ≤ N ≤ 200 000

1 ≤ M ≤ 200 000

1 ≤ a_i < b_i ≤ N

(a_i, b_i) \neq (1, N)

If i \neq j, (a_i, b_i) \neq (a_j, b_j).

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

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

POSSIBLE

Sample Input 2

4 3
1 2
2 3
3 4

Sample Output 2

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

100000 1
1 99999

Sample Output 3

IMPOSSIBLE

Sample Input 4

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

Sample Output 4

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.

Solution in Scala (2.11.7) (judge verdict: Accepted)

import java.util.Scanner

object Main extends App {
  val sc = new Scanner(System.in)

  val N, M = sc.nextInt()
  val graph = Array.fill[List[Int]](N+1)(Nil)
  for (_ <- 1 to M) {
    val a, b = sc.nextInt()
    graph(a) = b :: graph(a)
    graph(b) = a :: graph(b)
  }

  if(dfs(1, 0))
    println("POSSIBLE")
  else
    println("IMPOSSIBLE")

  def dfs(node: Int, cnt: Int): Boolean = {
    if (node == N)
      true
    else {
      if(cnt == 2)
        false
      else
        graph(node).map(dfs(_, cnt+1)).exists(identity)
    }
  }
}