#include <stdio.h>
#include <string.h>
/*我想用一个类似于端口插入的方法来解决这道题*/
/*每个点都有前后两个端口，可连接用 0 表示 已连接用 1 表示*/
int map[2000][2];

int main ()
{
     int m,n;
     int max = 0;
     int point = 0;
     int count = 0;

     while (scanf ("%d %d",&m,&n) != EOF)
     {
        if (m == -1 && n == -1)
        {
            break;
        }
        if (m == 0 && n == 0)
        {
            count++;
            /*排除森林的情况*/
            for (int i = 1;i < max;i++)
            {
                if (map[i][1] == 0 && map[i + 1][0] == 0)
                {
                    point = 1;
                    break;
                }
            }
            if (point == 1)
            printf ("Case %d is not a tree.\n",count);
            else
            printf ("Case %d is a tree.\n",count);

            for (int i = 0;i < max + 1;i++)
            {
                map[i][0] = 0;
                map[i][1] = 0;
            }
            point = 0;
            max = 0;
            continue;
        }
        /*排除插入同一个点左端口的情况*/
        if (map[n][0] == 1)
        {
            point = 1;
        }
        /*排除一个点已经有右端口牵出，插入左端口的情况*/
        else if (map[m][0] == 1 && map[n][1] == 1)
        {
            point = 1;
        }
        /*排除一个点左右端口自我连接的情况*/
        if(m == n)
        {
            point = 1;
        }
        else
        {
            map[n][0] = 1;
            map[m][1] = 1;
        }
        
        if (max < m)
        max = m;
        if (max < n)
        max = n;
     }
}