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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define lson (rt << 1)
#define rson (rt << 1 | 1)
using namespace std;
const int MAXN = 4e5 + 7;
int n, m; 
struct node{int v, next;}e[MAXN];
int head[MAXN], d[MAXN], tot = 1, vis[MAXN];
void add(int u, int v) {e[++tot] = {v, head[u]}; head[u] = tot; d[v]++;}
int a[MAXN << 1], cnt = 0;
void dfs(int u)
{
    for (int &i = head[u]; i; i = e[i].next)
    {
        if (vis[i]) continue;
        vis[i] = vis[i ^ 1] = 1;
        dfs(e[i].v);
    }
    a[++cnt] = u;
}
int pos[MAXN], nxt[MAXN];
int T[MAXN << 2];
inline void pushup(int rt) {T[rt] = min(T[lson], T[rson]);}
void build(int rt, int l, int r)
{
    if (l == r) {T[rt] = nxt[l]; return;}
    int mid = (l + r) >> 1;
    build(lson, l, mid); build(rson, mid + 1, r);
    pushup(rt);
}
int query(int rt, int l, int r, int L, int R)
{
    if (L <= l && r <= R) return T[rt];
    int res = 2e9, mid = (l + r) >> 1;
    if (L <= mid) res = min(res, query(lson, l, mid, L, R));
    if (mid <  R) res = min(res, query(rson, mid + 1, r, L, R));
    return res;
}
void print(int *a, int n)
{
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    puts("");
}
bool check()
{
    cnt--;
    for (int i = cnt + 1; i <= cnt * 2; i++) a[i] = a[i - cnt];
    for (int i = 1; i <= cnt * 2; i++) nxt[i] = 2e9;
    for (int i = 1; i <= cnt * 2; i++) nxt[pos[a[i]]] = i, pos[a[i]] = i;
    //print(nxt, cnt * 2);
    build(1, 1, cnt * 2);
    for (int i = 1; i <= cnt; i++) if (nxt[i] <= i + cnt - 1)
    {
        int pos = query(1, 1, cnt * 2, nxt[i], i + cnt - 1);
        if (pos <= i + cnt - 1) 
        {
            //printf("st = %d pos = %d\n", i, pos);
            return true;
        }
    }
    return false;
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1, u, v; i <= m; i++) scanf("%d%d", &u, &v), add(u, v), add(v, u);
    for (int i = 1; i <= n; i++) if (d[i] & 1)
    {
        puts("No");
        return 0;
    }
    dfs(1); 
    //print(a, cnt);
    check() ? puts("Yes") : puts("No");
    return 0;
}
/*
5 7
1 5
5 4
4 1
1 3
3 4
2 4
1 2
*/