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: Time Limit Exceeded)

/// Rustで競技プログラミングの入力をスッキリ記述するマクロ
///
/// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[inline]
#[allow(dead_code)]
fn gcd(mut m: u64, mut n: u64) -> u64 {
    if m < n {
        std::mem::swap(&mut m, &mut n);
    }

    loop {
        let k = m % n;
        if k == 0 {
            return n;
        }

        m = n;
        n = k;
    }
}

#[inline]
#[allow(dead_code)]
fn lcm(m: u64, n: u64) -> u64 {
    m / gcd(m, n) * n
}

macro_rules! eprint(
    ($($arg:tt)*) => { {
        use std::io::Write;
        let r = write!(&mut ::std::io::stderr(), $($arg)*);
        r.expect("failed printing to stderr");
    } }
);

macro_rules! eprintln(
    ($($arg:tt)*) => { {
        use std::io::Write;
        let r = writeln!(&mut ::std::io::stderr(), $($arg)*);
        r.expect("failed printing to stderr");
    } }
);

macro_rules! dbg {
    ($val:expr) => {
        // Use of `match` here is intentional because it affects the lifetimes
        // of temporaries - https://stackoverflow.com/a/48732525/1063961
        match $val {
            tmp => {
                eprintln!("[{}:{}] {} = {:#?}",
                          file!(), line!(), stringify!($val), &tmp);
                tmp
            }
        }
    }
}

fn solve(ab_v: &Vec<Vec<u64>>, n: u64) -> bool {
    for ab in ab_v {
        if ab[0] == 1 {
            let a = ab[1];
            for ab in ab_v {
                if a < n {
                    if ab[0] == a && ab[1] == n {
                        return true;
                    }
                } else {
                    if ab[0] == n && ab[1] == a {
                        return true;
                    }
                }
            }
        }
    }

    false
}

fn main() {
    input!{
        n: u64,
        m: usize,
        ab_v: [[u64; 2]; m],
    }

    println!("{}", if solve(&ab_v, n) {
        "POSSIBLE"
    } else {
        "IMPOSSIBLE"
    });
}