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.

大意：找出有多少个环

思路：算出每个顶点的度，只有度等于2的情况下才能成为环。如果两个顶点的度都为2，使用并查集，如果父节点相同，答案加一，否则联立。

AC代码：

#include<cstdio>
#include <map> 
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')f=-1,c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return f*x;}
typedef long long ll;
const int maxn=5e5+10;
int f[maxn];
int sum[maxn];
int ans;
struct node{
    int u,v;
}a[maxn];
int find(int x){
    if(f[x]==x){
        return x;
    }
    else{
        return f[x]=find(f[x]);
    }
}
void unio(int x,int y){
    int f1=find(x);
    int f2=find(y);
    if(f1!=f2){
        f[f1]=f2;
    }
    else
        ans++;
}
int main(){
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        f[i]=i;
    }
    ans=0;
    for(int i=0;i<m;i++){
        cin>>a[i].u>>a[i].v;
        sum[a[i].u]++;
        sum[a[i].v]++;
    }
    for(int i=0;i<m;i++){
        if(sum[a[i].u]==2&&sum[a[i].v]==2){
            unio(a[i].u,a[i].v);
        }
    }
    printf("%d",ans);
}