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::vec::Vec;
use std::cmp::*;
use std::collections::BinaryHeap;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
  let (n, m) = {
    let v: Vec<usize> = read_vec();
    (v[0], v[1])
  };

  let mut g: Vec<Vec<((usize, u64))>> = vec![vec![]; n];
  for _ in 0..m {
    let t: Vec<usize> = read_vec();
    g[t[0] - 1].push((t[1] - 1, 1));
  }

  let d = dijkstra_to_goal(n, &g, 0, n-1);
  if d == 2 {
    println!("POSSIBLE");
  } else {
    println!("IMPOSSIBLE");
  }
}



// 単一始点から全てのノードへの最短経路超を探索
fn dijkstra_to_goal(n: usize, adj_list: &Vec<Vec<(usize, u64)>>, start: usize, goal: usize) -> u64 {
  let mut dist = vec![std::u64::MAX; n];
  let mut p_queue: BinaryHeap<(i64, usize)> = BinaryHeap::new();

  //init
  dist[start] = 0;
  p_queue.push((0, start));

  while !p_queue.is_empty() {
    // 起点となるノード
    let (cost, now)
      = match p_queue.pop() {
        None => break,
        Some(e) => (-e.0 as u64, e.1),
      };

      if now == goal {
        break;
      }

    // 繋がっているノードを探索
    for &edge in &adj_list[now] {
      let next = edge.0;
      let weight = edge.1;

      if dist[next] > weight + cost {
        p_queue.push((-((weight + cost) as i64), next));
        dist[next] = weight + cost;
      }
    }

  }

  dist[goal]
}