NIO is playing a game about trees.

The game has two trees each with vertices. The vertices in each tree are numbered from to and the -th vertex has the weight . The root of each tree is vertex 1. Given key numbers , find the number of solutions that remove exactly one number so that the weight of the lowest common ancestor of the vertices in A with the remaining key numbers is greater than the weight of the lowest common ancestor of the vertices in B with the remaining key numbers.

输入描述:

The first line has two positive integers .

The second line has  unique positive integers .

The third line has  positive integers  represents the weight of vertices in A.

The fourth line has  positive integers , indicating that the number of the father of vertices  in tree A is .

The fifth line has  positive integers  represents the weight of vertices in B.

The sixth line has  positive integers , indicating that the number of the father of vertices  in tree B is .

输出描述:

One integer indicating the answer.

5 3
5 4 3
6 6 3 4 6
1 2 2 4
7 4 5 7 7
1 1 3 2

1

In first case, the key numbers are 5,4,3. 
Remove key number 5, the lowest common ancestors of the vertices in A with the remaining key numbers is 2, in B is 3.
Remove key number 4, the lowest common ancestors of the vertices in A with the remaining key numbers is 2, in B is 1.
Remove key number 3, the lowest common ancestors of the vertices in A with the remaining key numbers is 4, in B is 1.
Only remove key number 5 satisfies the requirement.

10 3
10 9 8
8 9 9 2 7 9 0 0 7 4
1 1 2 4 3 4 2 4 7
7 7 2 3 4 5 6 1 5 3
1 1 3 1 2 4 7 3 5

2

备注:

The lowest common ancestor (LCA) (also called least common ancestor) of two nodes v and w in a tree or directed acyclic graph (DAG) T is the lowest (i.e. deepest) node that has both v and w as descendants, where we define each node to be a descendant of itself (so if v has a direct connection from w, w is the lowest common ancestor).(From Wiki.)