Problem:
Score : 800 points

Problem Statement

You are given a simple connected undirected graph consisting of N vertices and M edges.
The vertices are numbered 1 to N, and the edges are numbered 1 to M.

Edge i connects Vertex a_i and b_i bidirectionally.

Determine if three circuits (see Notes) can be formed using each of the edges exactly once.

Notes

A circuit is a cycle allowing repetitions of vertices but not edges.

Constraints

All values in input are integers.

1 \leq N,M \leq 10^{5}

1 \leq a_i, b_i \leq N

The given graph is simple and connected.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_M b_M

Output

If three circuits can be formed using each of the edges exactly once, print Yes; if they cannot, print No.

Sample Input 1

7 9
1 2
1 3
2 3
1 4
1 5
4 5
1 6
1 7
6 7

Sample Output 1

Yes

Three circuits can be formed using each of the edges exactly once, as follows:

Sample Input 2

3 3
1 2
2 3
3 1

Sample Output 2

No

Three circuits are needed.

Sample Input 3

18 27
17 7
12 15
18 17
13 18
13 6
5 7
7 1
14 5
15 11
7 6
1 9
5 4
18 16
4 6
7 2
7 11
6 3
12 14
5 2
10 5
7 8
10 15
3 15
9 8
7 15
5 16
18 15

Sample Output 3

Yes

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

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)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($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, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (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")
    };
}

use std::collections::{HashSet, VecDeque};

fn main() {
    input!{
        n: usize,
        m: usize,
        v: [(usize1, usize1); m]
    }
    let mut g: Vec<HashSet<usize>> = vec![HashSet::new(); n];
    for &(a, b) in v.iter() {
        g[a].insert(b);
        g[b].insert(a);
    }
    for i in 0..n {
        if g[i].len() % 2 != 0 {
            println!("No");
            return;
        }
    }
    let ans = bfs(&g, &(0..n).collect::<HashSet<usize>>(), &HashSet::new(), 0);
    println!("{}", if ans {"Yes"} else {"No"});
}

fn bfs(g: &Vec<HashSet<usize>>, setv: &HashSet<usize>, sete: &HashSet<(usize, usize)>, k: usize) -> bool {
    if k >= 3 {
        return true;
    }
    if setv.len() == 0 {
        return false;
    }
    let mut q: VecDeque<(usize, HashSet<usize>, HashSet<(usize, usize)>)> = VecDeque::new();
    let mut setv2 = setv.clone();
    let mut n0 = 0;
    while !setv2.contains(&n0) {
        n0 += 1;
    }
    setv2.remove(&n0);
    q.push_back((n0, setv2, sete.clone()));
    while q.len() > 0 {
        let (n, mut sv, mut se) = q.pop_front().unwrap();
        for &c in g[n].iter() {
            if se.contains(&(n, c)) || se.contains(&(c, n)) {
                continue;
            }
            se.insert((n, c));
            if !sv.contains(&c) {
                return bfs(g, &sv, &se, k + 1);
            }
            sv.remove(&c);
            q.push_back((c, sv.clone(), se.clone()));
        }
    }
    return false;
}