Problem description

A graph consists of a set of vertices and edges between pairs of vertices. Two vertices are connected if there is a path (subset of edges) leading from one vertex to another, and a connected component is a maximal subset of vertices that are all connected to each other. A graph consists of one or more connected components.

A tree is a connected component without cycles, but it can also be characterized in other ways. For example, a tree consisting of n vertices has exactly n-1 edges. Also, there is a unique path connecting any pair of vertices in a tree.

Given a graph, report the number of connected components that are also trees.

Input

The input consists of a number of cases. Each case starts with two non-negative integers n and m, satisfying n ≤ 500 and m ≤ n(n-1)/2. This is followed by m lines, each containing two integers specifying the two distinct vertices connected by an edge. No edge will be specified twice (or given again in a different order). The vertices are labelled 1 to n. The end of input is indicated by a line containing n = m = 0.

Output

For each case, print one of the following lines depending on how many different connected components are trees (T > 1 below):

Case x: A forest of T trees.	Case x: There is one tree.	Case x: No trees.

x is the case number (starting from 1).

Sample Input

6 31 22 33 46 51 22 33 44 55 66 61 22 31 34 55 66 40 0

Sample Output

Case 1: A forest of 3 trees.Case 2: There is one tree.Case 3: No trees.

#include
        
         int fath[505],cycl[505],k,n;void setfirst(){    k=n;    for(int i=1;i<=n;i++)    {        fath[i]=i; cycl[i]=0;    }}int find_fath(int x){    if(x!=fath[x])    fath[x]=find_fath(fath[x]);    return fath[x];}void setTree(int a,int b){    a=find_fath(a);    b=find_fath(b);    if(cycl[b]&&cycl[a])    return ;    k--;    if(a!=b)    {        if(cycl[a])        fath[b]=a;        else        fath[a]=b;    }    else    cycl[a]=1;}int main(){    int a,b,m,t=1;    while(scanf("%d%d",&n,&m)>0&&m+n!=0)    {        setfirst();        while(m--)        {            scanf("%d%d",&a,&b);            setTree(a,b);        }        printf("Case %d: ",t++);        if(k>1)printf("A forest of %d trees.\n",k);        if(k==1)printf("There is one tree.\n");        if(k==0)printf("No trees.\n");    }}