You are given an undirected graph consisting of $\mathrm{nn vertices\; and mm edges.\; Your\; task\; is\; to\; find\; the\; number\; of\; connected\; components\; which\; are\; cycles.}$

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 $\mathrm{aa is\; connected\; with\; a\; vertex bb,\; a\; vertex bb is\; also\; connected\; with\; a\; vertex aa).\; An\; edge\; can\text{'}t\; connect\; vertex\; with\; itself,\; there\; is\; at\; most\; one\; edge\; between\; a\; pair\; of\; vertices.}$

Two vertices $\mathrm{uu and vv belong\; to\; the\; same\; connected\; component\; if\; and\; only\; if\; there\; is\; at\; least\; one\; path\; along\; edges\; connecting uu and vv.}$

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 $\mathrm{6 connected\; components, 2 of\; them\; are\; cycles: [7,10,16]and [5,11,9,15].}$

Input

The first line contains two integer numbers $\mathrm{nn and mm (1\le n\le 2\cdot 10^5, 0\le m\le 2\cdot 10^5)\; —\; number\; of\; vertices\; and\; edges.}$

The following $\mathrm{mm lines\; contains\; edges:\; edge ii is\; given\; as\; a\; pair\; of\; vertices vi, ui (1\le vi,ui\le n, ui\ne vi).\; There\; is\; no\; multiple\; edges\; in\; the\; given\; graph,\; i.e.\; for\; each\; pair\; (vi,ui)\; there\; no\; other\; pairs\; (vi,ui)\; and\; (vi,ui)\; in\; the\; list\; of\; edges.}$

Output

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

Examples

input

Copy

5 4 1 2 3 4 5 4 3 5

output

Copy

1

input

Copy

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

Copy

2

Note

In the first example only component $[3,4,5] is\; also\; a\; cycle.$

The illustration above corresponds to the second example.

分析：DFS 如果其中一个连通图的所有点的度数都为2就符合题意(搜索完某一连通图就把该连通图的所有点做标记，不再访问)

代码：

#include
     
      using namespace std;const int N = 200000 + 5; vector
      
        a[N];int vis[N];int flag = 1;void dfs(int cur) {    vis[cur] = 1;    if(a[cur].size() != 2) flag = 0;    for(int i : a[cur]) {        if(!vis[i]) dfs(i);    }} int main() {    int n, m;    scanf("%d%d", &n, &m);    int x, y;    for(int i = 0; i < m; i++) {        scanf("%d%d", &x, &y);        a[x].push_back(y);        a[y].push_back(x);    }    memset(vis, 0, sizeof(vis));    int ans = 0;    for(int i = 1; i <= n; i++) {        flag = 1;            if(!vis[i]) {            dfs(i);            if(flag) ans++;        }    }    printf("%d\n", ans);    return 0;}