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: Accepted)

use std::io::BufRead;

fn main() {
    let s = std::io::stdin();
    let mut buf = s.lock();
    let mut s = String::new();
    let (n, m) = {
        buf.read_line(&mut s).unwrap();
        let mut it = s.split_whitespace();
        let n: usize = it.next().unwrap().parse().unwrap();
        let m: usize = it.next().unwrap().parse().unwrap();
        (n, m)
    };
    let mut a = vec![(false, false); n];
    for _ in 0..m {
        s.clear();
        buf.read_line(&mut s).unwrap();
        let mut it = s.split_whitespace();
        let x: usize = it.next().unwrap().parse().unwrap();
        let y: usize = it.next().unwrap().parse().unwrap();
        if x == 1 {
            a[y - 1].0 = true;
        }
        if y == n {
            a[x - 1].1 = true;
        }
    }
    for (a, b) in a {
        if a && b {
            println!("POSSIBLE");
            return;
        }
    }
    println!("IMPOSSIBLE");
}