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::*;
use std::str::*;

fn read_usize(s: &mut StdinLock) -> usize {
    s.by_ref().bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .fold(0, |a, x| (x as u8 - b'0') as usize + a * 10)
}

fn main(){
    let s = stdin();
    let mut s = s.lock();

    let n: usize = read_usize(&mut s);
    let m: usize = read_usize(&mut s);
    let mut from1 = vec![false; n + 1];
    let mut fromn = vec![false; n + 1];
    for _ in 0..m {
        let a: usize = read_usize(&mut s);
        let b: usize = read_usize(&mut s);
        from1[b] |= a == 1;
        from1[a] |= b == 1;
        fromn[b] |= a == n;
        fromn[a] |= b == n;
    }
    let flg = from1.iter().zip(fromn.iter()).fold(false, |acc, (&a, &b)| acc || a && b);
    println!("{}", if flg { "POSSIBLE" } else { "IMPOSSIBLE" });
}