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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.*;
import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int m = sc.nextInt();

    ArrayList<Integer>[] adj = new ArrayList[n];
    for(int i=0;i<n;i++) adj[i] = new ArrayList<Integer>();

    for(int i=0;i<m;i++) {
      int a = sc.nextInt() - 1;
      int b = sc.nextInt() - 1;
      adj[a].add(b);
      adj[b].add(a);
    }

    Queue<Integer> q = new LinkedList<>();
    q.add(1);
    int cnt = 0;

    boolean[] vis = new boolean[n];

    bfs:
    while(!q.isEmpty()) {
        int cur = q.poll();
        for(int next : adj[cur]) {
          if(next == n - 1) {
            break bfs;
          }
          if(!vis[next]) {
            cnt++;
          }
        }
    }

    System.out.println(cnt <= 2 ? "POSSIBLE" : "IMPOSSIBLE");
  }
}