Coloring a Tree（耐心翻译+思维）

Description

You are given a rooted tree with n vertices. The vertices are numbered from 1 to n, the root is the vertex number 1.

Each vertex has a color, let's denote the color of vertex v by c_v. Initially c_v = 0.

You have to color the tree into the given colors using the smallest possible number of steps. On each step you can choose a vertex v and a color x, and then color all vectices in the subtree of v (including v itself) in color x. In other words, for every vertex u, such that the path from root to u passes through v, set c_u = x.

It is guaranteed that you have to color each vertex in a color different from 0.

You can learn what a rooted tree is using the link: https://en.wikipedia.org/wiki/Tree_(graph_theory).

Input

The first line contains a single integer n (2 ≤ n ≤ 10⁴) — the number of vertices in the tree.

The second line contains n - 1 integers p₂, p₃, ..., p_n (1 ≤ p_i < i), where p_i means that there is an edge between vertices i and p_i.

The third line contains n integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ n), where c_i is the color you should color the i-th vertex into.

It is guaranteed that the given graph is a tree.

Output

Print a single integer — the minimum number of steps you have to perform to color the tree into given colors.

Sample Input

Input

6
1 2 2 1 5
2 1 1 1 1 1

Output

3

Input

7
1 1 2 3 1 4
3 3 1 1 1 2 3

Output

5

Hint

The tree from the first sample is shown on the picture (numbers are vetices' indices):

On first step we color all vertices in the subtree of vertex 1 into color 2 (numbers are colors):

On seond step we color all vertices in the subtree of vertex 5 into color 1:

On third step we color all vertices in the subtree of vertex 2 into color 1:

The tree from the second sample is shown on the picture (numbers are vetices' indices):

On first step we color all vertices in the subtree of vertex 1 into color 3 (numbers are colors):

On second step we color all vertices in the subtree of vertex 3 into color 1:

On third step we color all vertices in the subtree of vertex 6 into color 2:

On fourth step we color all vertices in the subtree of vertex 4 into color 1:

On fith step we color all vertices in the subtree of vertex 7 into color 3:

   题目意思： 给你n个点，代表这一棵树有n个节点。第二行内容是建树的关系，（开始我一直看不明白是如何建树的，后来翻译给队友，他帮我指了）

从第二个节点开始的节点和父节点（上一个节点）相连，

例如：1 2 2 1 5
代表：节点2和节点1相连，节点3和节点2相连，节点4和节点2相连，节点5和节点1相连，节点6和节点5相连。

第三行内容是需要将各个点涂成的颜色，给这个树涂色，有这么一条原则就是给某一节点涂色，以其为根节点的子树也将变为相应的颜色，我们可以成为一种颜料的溢出，问你最终需要
最少需要涂多少次颜色就可以满足题目要求。

解题思路：我们可以这样来思考，因为最后需要使所有的点都涂成要求的颜色，一定是按照从根节点到叶子节点遍历的涂色，但所有的点都遍历会造成浪费，我们只需要找出需要涂的点即可，
那么哪些点需要涂呢？我们发现只有那些最后要求的其父亲节点和本身不同色的需要涂色，因为需要向下改变自身颜色，那么只需要统计这样点的个数即可。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int pr[10010];
 int a[10010];
 int main()
 {
     int n,i,counts;
     scanf("%d",&n);
     counts=0;
     pr[0]=1;
     pr[1]=1;///根节点的父亲节点是自身
     for(i=2;i<=n;i++)
     {
         scanf("%d",&pr[i]);
     }
     for(i=1;i<=n;i++)
     {
         scanf("%d",&a[i]);
     }
     for(i=1;i<=n;i++)
     {
         if(a[i]!=a[pr[i]])///父亲节点和自身颜色不同
         {
             counts++;
         }
     }
     printf("%d\n",counts);
     return 0;
 }