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, Write};
use std::io;

fn solve<R: BufRead, W: Write>(mut input: R, mut output: W) {
    let mut s = String::new();
    input.read_line(&mut s).unwrap();
    let mut split = s.split_whitespace();
    let n: usize = split.next().unwrap().parse().unwrap();
    let m: usize = split.next().unwrap().parse().unwrap();
    let mut adjacency = vec![Vec::new(); n];
    for _ in 0..m {
        let mut s = String::new();
        input.read_line(&mut s).unwrap();
        let mut split = s.split_whitespace();
        let a: usize = split.next().unwrap().parse().unwrap();
        let b: usize = split.next().unwrap().parse().unwrap();
        adjacency[a - 1].push(b - 1);
        adjacency[b - 1].push(a - 1);
    }
    for &i in &adjacency[0] {
        for &j in &adjacency[i] {
            if j == n - 1 {
                return writeln!(output, "POSSIBLE").unwrap();
            }
        }
    }
    writeln!(output, "IMPOSSIBLE").unwrap();
}

fn main() {
    let stdin = io::stdin();
    let stdout = io::stdout();
    solve(stdin.lock(), stdout.lock());
}

#[cfg(test)]
mod tests {
    fn test(input: &str, output: &str) {
        let mut v: Vec<u8> = Vec::new();
        ::solve(input.as_bytes(), &mut v);
        assert_eq!(String::from_utf8(v).unwrap(), output);
    }
    #[test]
    fn case1() {
        test(
            "3 2
1 2
2 3\n",
            "POSSIBLE\n",
        );
    }
    #[test]
    fn case2() {
        test(
            "4 3
1 2
2 3
3 4\n",
            "IMPOSSIBLE\n",
        );
    }
    #[test]
    fn case3() {
        test(
            "100000 1
1 99999\n",
            "IMPOSSIBLE\n",
        );
    }
    #[test]
    fn case4() {
        test(
            "5 5
1 3
4 5
2 3
2 4
1 4\n",
            "POSSIBLE\n",
        );
    }
}