Cyclic Components

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 $\mathrm{66\; connected\; components,\; 22\; of\; them\; are\; cycles:\; [7,10,16][7,10,16]\; and\; [5,11,9,15][5,11,9,15]\; .}$

Input

The first line contains two integer numbers $\mathrm{nn\; and\; mm\; (1\le n\le 2\cdot 1051\le n\le 2\cdot 105\; ,\; 0\le m\le 2\cdot 1050\le m\le 2\cdot 105\; )\; —\; number\; of\; vertices\; and\; edges.}$

The following $\mathrm{mm\; lines\; contains\; edges:\; edge\; ii\; is\; given\; as\; a\; pair\; of\; vertices\; vivi\; ,\; uiui\; (1\le vi,ui\le n1\le vi,ui\le n\; ,\; ui\ne viui\ne vi\; ).\; There\; is\; no\; multiple\; edges\; in\; the\; given\; graph,\; i.e.\; for\; each\; pair\; (vi,uivi,ui\; )\; there\; no\; other\; pairs\; (vi,uivi,ui\; )\; and\; (ui,viui,vi\; )\; in\; the\; list\; of\; edges.}$

Output

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

Examples

Input

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 $[3,4,5][3,4,5]\; is\; also\; a\; cycle.$

The illustration above corresponds to the second example.

题解：用并查集加度数都为2的边，出现一次矛盾ans即加1.因为环内的点度数都为。

#pragma warning(disable:4996)
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long 

const int maxn = 200005;

int n, m, cnt;
int x[maxn], y[maxn], pa[maxn];

struct node{
    int u,v;
}e[maxn];

int Find(int a) {
    if (a == pa[a]) return a;
    return pa[a] = Find(pa[a]);
}

void Union(int a, int b) {
    int tx = Find(a);
    int ty = Find(b);
    if (tx == ty) cnt++;
    else pa[tx] = ty;
}

void Inite()
{
    memset(x, 0, sizeof(x));
    for (int i = 1; i <= n; i++) pa[i] = i;
}


int main()
{
    while (scanf("%d%d", &n, &m) != EOF) {
        Inite();
        for (int i = 1; i <= m; i++) {
            scanf("%d%d", &e[i].u, &e[i].v);
            if (e[i].u == e[i].v) continue;
            x[e[i].u]++;
            x[e[i].v]++;
        }
        cnt = 0;
        for (int i = 1; i <= m; i++) {
            if (x[e[i].u] == 2 && x[e[i].v] == 2) {
                Union(e[i].u, e[i].v);
            }
        }
        cout << cnt << endl;
    }
    return 0;
}