E. Cyclic Components

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You are given an undirected graph consisting of n vertices and m

edges. Your task is to find the number of connected components which are cycles.

Here are some definitions of graph theory.

An undirected graph consists of two sets: set of nodes (called vertices) and set of edges. Each edge connects a pair of vertices. All edges are bidirectional (i.e. if a vertex a

is connected with a vertex b, a vertex b is also connected with a vertex a

). An edge can't connect vertex with itself, there is at most one edge between a pair of vertices.

Two vertices u

and v belong to the same connected component if and only if there is at least one path along edges connecting u and v

.

A connected component is a cycle if and only if its vertices can be reordered in such a way that:

• the first vertex is connected with the second vertex by an edge,
• the second vertex is connected with the third vertex by an edge,
• ...
• the last vertex is connected with the first vertex by an edge,
• all the described edges of a cycle are distinct.

A cycle doesn't contain any other edges except described above. By definition any cycle contains three or more vertices.

There are 6 connected components, 2 of them are cycles: [7,10,16] and [5,11,9,15]

. Input

The first line contains two integer numbers n

and m ( 1≤n≤2⋅105, 0≤m≤2⋅105

) — number of vertices and edges.

The following m

lines contains edges: edge i is given as a pair of vertices vi, ui ( 1≤vi,ui≤n, ui≠vi). There is no multiple edges in the given graph, i.e. for each pair ( vi,ui) there no other pairs ( vi,ui) and ( ui,vi

) in the list of edges.

Output

Print one integer — the number of connected components which are also cycles.

Examples

Input

5 4
1 2
3 4
5 4
3 5

Output

1

Input

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

Output

2

Note

In the first example only component [3,4,5]

is also a cycle.

The illustration above corresponds to the second example.

题意：给你n个点，m条边，问其中环的个数

分析：1、环的定义是每个点的度为二，也就是不能有多余的分支

            2、dfs找到同一个集合的点，然后判断所有点的度是否为2，来判断是否成环

           3、判断点的度可以用邻接表中每个点对应的大小

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#include <vector>
vector<int> g[200010];
vector<int> di;
int vis[200010]={0};
void dfs(int x)  //找到同一个集合的点
{
   vis[x]=1;     //避免重复查找
   di.push_back(x);    //存储同一个集合的点
   for(int i=0;i<g[x].size();i++)  //遍历对应的邻接表
   {
      if(vis[g[x][i]]==0)
        dfs(g[x][i]);
   }
}
int main()
{
   int n,m,x,y,i,j,cnt=0;
   scanf("%d%d",&n,&m);
   for(i=0;i<m;i++)
   {
      scanf("%d%d",&x,&y);
      g[x].push_back(y);   //存储邻接表
      g[y].push_back(x);
   }
   for(i=1;i<=n;i++)
   {
      if(!vis[i])
      {
         di.clear();   //清空存储仓
         dfs(i);
         for(j=0;j<di.size();j++)   //判断是否成环，所有点的度均为2
         {
            if(g[di[j]].size()!=2)
              break;
         }
         if(j==di.size())
           cnt++;
      }
   }
   printf("%d\n",cnt);
   return 0;
}