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: Memory Limit Exceeded)

import java.util.Scanner;
public class Main {
	static boolean g[][];
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.next());
		int m = Integer.parseInt(sc.next());
		g = new boolean[n][n];
		for (int i = 0; i < m; i++) {
			int a = Integer.parseInt(sc.next()) - 1;
			int b = Integer.parseInt(sc.next()) - 1;
			g[a][b] = g[b][a] = true;
		}
		boolean[] visited = new boolean[n];
		visited[0] = true;
		int ans = dfs(0,n,visited);
		if (ans > 0)
			System.out.println("POSSIBLE");
		else
			System.out.println("IMPOSSIBLE");
	}
	public static int dfs(int v, int n, boolean[] visited) {
		if (visited[n - 1]) {
			int count = 0;
			for (int i = 1; i < n - 1; i++)
				if (visited[i])
					count++;
			if (count == 1)
				return 1;
		}
		int ans = 0;
		for (int i = 0; i < n; i++) {
			if (g[v][i] == false) continue;
			if (visited[i]) continue;
			visited[i] = true;
			ans += dfs(i, n, visited);
			visited[i] = false;
		}
		return ans;
	}
}