Problem Description

You are given an undirected graph consisting of nn vertices and mm 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 aa is connected with a vertex bb, a vertex bb is also connected with a vertex aa). An edge can't connect vertex with itself, there is at most one edge between a pair of vertices.

Two vertices uu and vv belong to the same connected component if and only if there is at least one path along edges connecting uu and vv.

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 66 connected components, 22 of them are cycles: [7,10,16][7,10,16] and [5,11,9,15][5,11,9,15].

Input

The first line contains two integer numbers nn and mm (1≤n≤2⋅1051≤n≤2⋅105, 0≤m≤2⋅1050≤m≤2⋅105) — number of vertices and edges.

The following mm lines contains edges: edge ii is given as a pair of vertices vivi, uiui (1≤vi,ui≤n1≤vi,ui≤n, ui≠viui≠vi). There is no multiple edges in the given graph, i.e. for each pair (vi,uivi,ui) there no other pairs (vi,uivi,ui) and (ui,viui,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][3,4,5] is also a cycle.

The illustration above corresponds to the second example.

 

本题是图上DFS。根据题意。每个在圈上的点的入度和出度都为2（因为是双向图）。因此如果不是2的话就令与该点相关的点都无效即可。

（本题也可以用并查集。存所有度为2的点。判断是否在一个子图（并查集还不太会，以后再说qwq））

AC代码：

#include <algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#define MAXN 200005
using namespace std;
int n,m,temp1,temp2,cnt,vis[MAXN],flag=1;
vector<int>g[MAXN];
int dfs(int x){
    vis[x]=1;
    if(g[x].size()!=2)
        flag = 0;//如果该点的度不为2.则继续。不过此时flag已经为0.故是排除选项 
    for(int i=0;i<g[x].size();i++){
        if(!vis[g[x][i]])
            dfs(g[x][i]);
        }
    return 0;    
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++){
        scanf("%d%d",&temp1,&temp2);
        g[temp1].push_back(temp2);//双向边 
        g[temp2].push_back(temp1);
    }
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++){
        if(vis[i])
            continue;
        dfs(i);
        if(flag)
            cnt++;
        else
            flag=1;//重置 
    }    
    printf("%d\n",cnt);
       return 0;
}