#11175From D to E and Back

Anyone who goes to a psychiatrist ought to have his head examined. Samuel Goldwyn Take any directed graph D with n vertices and m edges. You can make the Lying graph E of B in the following way. E will have m vertices, one for each edge of D. For example, if D has an edge uv, then E will have a vertex called uv. Now, whenever D has edges uv and vw, E will have an edge from vertex uv to vertex vw. There are no other edges in E. You will be given a graph E and will have to determine whether it is possible for E to be the Lying graph of some directed graph D.

Input

The first line of input gives the number of cases, N (N < 220). N test cases follow. Each one starts with two lines containing m (0 ≤ m ≤ 300) and k. The next k lines will each contain a pair of vertices, x and y, meaning that there is an edge from x to y in E. The vertices are numbered from 0 to m − 1

Output

For each test case, output one line containing ‘Case #x:’ followed by either ‘Yes’ or ‘No’, depending on whether E is a valid Lying graph or not. Note that D is allowed to have duplicate edges and self-edges.

4
2
1
0 1
5
0
4
3
0 1
2 1
2 3
3
9
0 1
0 2
1 2
1 0
2 0
2 1
0 0
1 1
2 2

Case #1: Yes
Case #2: Yes
Case #3: No
Case #4: Yes