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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
    "fmt"
)

func main() {
    var n, m int 
    fmt.Scan(&n, &m) 
    ns := make([]int, n)
    for i := 0; i < m; i++ {
        var a, b int 
        fmt.Scan(&a, &b) 
        ns[a-1]++
        ns[b-1]++
    }   

    var ng bool
    var odd, two, four, six int 
    for _, v := range ns {
        if v < 2 { 
            ng = true
        }   
        if v%2 == 1 { 
            odd++
        }   
        if v >= 2 { 
            two++
        }   
        if v >= 4 { 
            four++
        }   
        if v >= 6 { 
            six++
        }   
    }   

    if ng {
        fmt.Println("No")
    } else {
        switch odd {
        case 0:
            if six >= 1 || four >= 2 { 
                fmt.Println("Yes")
            } else {
                fmt.Println("No")
            }   
        case 2:
            if four >= 1 { 
                fmt.Println("Yes")
            } else {
                fmt.Println("No")
            }   
        case 4:
            if two >= 2 { 
                fmt.Println("Yes")
            } else {
                fmt.Println("No")
            }   
        default:
            fmt.Println("No")
        }   
    }   
}