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: Time Limit Exceeded)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		sc.nextLine();
		int[] arrayA = new int[m];
		int[] arrayB = new int[m];
		for (int i = 0; i < m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			if (a == 1) {
//				if (listB.contains(b)) {
//					System.out.println("POSSIBLE");
//					return;
//				}
				Arrays.fill(arrayA, b);
			} else if (a == n) {
				Arrays.fill(arrayB, b);
			}
			if (b == 1) {
//				if (listA.contains(a)) {
//					System.out.println("POSSIBLE");
//					return;
//				}
				Arrays.fill(arrayA, a);
			} else if (b == n) {
				Arrays.fill(arrayB, a);
			}
			sc.nextLine();
		}
		sc.close();
		String result = "IMPOSSIBLE";
		for (int i : arrayA) {
			if (Arrays.binarySearch(arrayB, i) >= 0) {
				result = "POSSIBLE";
				break;
			}
		}
		System.out.println(result);
	}
}