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 Rust (1.15.1) (judge verdict: Runtime Error)

fn main(){
	let mut s = String::new();
	std::io::stdin().read_line(&mut s).unwrap();
	let vec: Vec<&str> = s.trim().split(' ').collect();
	let n: i32 = vec[0].parse().unwrap();
	let m: i32 = vec[1].parse().unwrap();
	let mut a: Vec<i32> = Vec::new();
	let mut b: Vec<i32> = Vec::new();
	for i in 0..n {
		let mut d = String::new();
		std::io::stdin().read_line(&mut d).unwrap();
		let vec: Vec<&str> = d.trim().split(' ').collect();
		a.push(vec[0].parse().unwrap());
		b.push(vec[1].parse().unwrap());
	}
	let mut last = 0;
	let mut last_no: usize = 0;
	for i in 0..b.len() {
		if b[i] == m {
			last = b[i];
			last_no = i;
		}
		for j in 0..a.len() {
			if b[j] == last {
				if last_no == i {
					println!("POSSIBLE");
					break;
				}
			} else {
				println!("IMPOSSIBLE");
				break;
			}
		}
	}
}