You have a root tree containing n vertexes. Let's number the tree vertexes with integers from 1 to n. The tree root is in the vertex 1.

Each vertex (except fot the tree root) v has a direct ancestor pv. Also each vertex v has its integer value sv. 

Your task is to perform following queries:

  * P v u (u ≠ v). If u isn't in subtree of v, you must perform the assignment pv = u. Otherwise you must perform assignment pu = v. Note that after this query the graph continues to be a tree consisting of n vertexes.
  * V v t. Perform assignment sv = t. 



Your task is following. Before starting performing queries and after each query you have to calculate expected value written on the lowest common ancestor of two equiprobably selected vertices i and j. Here lowest common ancestor of i and j is the deepest vertex that lies on the both of the path from the root to vertex i and the path from the root to vertex j. Please note that the vertices i and j can be the same (in this case their lowest common ancestor coincides with them).

Input

The first line of the input contains integer n (2 ≤ n ≤ 5·104) — the number of the tree vertexes. 

The second line contains n - 1 integer p2, p3, ..., pn (1 ≤ pi ≤ n) — the description of the tree edges. It is guaranteed that those numbers form a tree.

The third line contains n integers — s1, s2, ... sn (0 ≤ si ≤ 106) — the values written on each vertex of the tree.

The next line contains integer q (1 ≤ q ≤ 5·104) — the number of queries. Each of the following q lines contains the description of the query in the format described in the statement. It is guaranteed that query arguments u and v lie between 1 and n. It is guaranteed that argument t in the queries of type V meets limits 0 ≤ t ≤ 106.

Output

Print q + 1 number — the corresponding expected values. Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 9.

Examples

Input

5
1 2 2 1
1 2 3 4 5
5
P 3 4
P 4 5
V 2 3
P 5 2
P 1 4


Output

1.640000000
1.800000000
2.280000000
2.320000000
2.800000000
1.840000000

Note

Note that in the query P v u if u lies in subtree of v you must perform assignment pu = v. An example of such case is the last query in the sample.
#include <bits/stdc++.h>
using namespace std;
const int N = 100100;
const int mod = 998244353;
inline int read(int x = 0, bool f = 0, char ch = getchar()) {
  for (; !isdigit(ch); ch = getchar()) f = ch == '-';
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
  return f ? -x : x;
}
int n, m, fa[N], ch[N][2], p[N];
long long f[N], siz[N], res[N], ans[N], ssiz[N], sum[N], a[N];
vector<int> g[N];
inline bool get(int x) { return ch[fa[x]][1] == x; }
inline bool gr(int x) { return ch[fa[x]][0] == x || ch[fa[x]][1] == x; }
inline void pushup(int x) {
  ssiz[x] = siz[x] + ssiz[ch[x][0]] + ssiz[ch[x][1]];
  sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + siz[x] * a[x];
  ans[x] = ans[ch[x][0]] + ans[ch[x][1]] + (siz[x] * siz[x] + f[x]) * a[x] +
           2 * siz[x] * ssiz[ch[x][1]] * a[x] +
           2 * sum[ch[x][0]] * (ssiz[x] - ssiz[ch[x][0]]) + res[x];
}
inline void rotate(int x) {
  int y = fa[x], z = fa[y], k = get(x), w = ch[x][k ^ 1];
  if (gr(y)) ch[z][get(y)] = x;
  fa[x] = z;
  ch[x][k ^ 1] = y;
  fa[y] = x;
  ch[y][k] = w;
  fa[w] = y;
  pushup(y);
  pushup(x);
}
inline void splay(int x) {
  for (; gr(x); rotate(x))
    if (gr(fa[x])) rotate(get(fa[x]) == get(x) ? fa[x] : x);
}
inline void add(int x, int y) {
  siz[x] += ssiz[y];
  f[x] -= ssiz[y] * ssiz[y];
  res[x] += ans[y];
  pushup(x);
}
inline void del(int x, int y) {
  siz[x] -= ssiz[y];
  f[x] += ssiz[y] * ssiz[y];
  res[x] -= ans[y];
  pushup(x);
}
inline void access(int x) {
  for (int y = 0; x; y = x, x = fa[x]) {
    splay(x);
    del(x, y);
    add(x, ch[x][1]);
    ch[x][1] = y;
    pushup(x);
  }
}
inline void link(int x, int y) {
  access(x);
  splay(x);
  splay(y);
  fa[y] = x;
  add(x, y);
  access(y);
  splay(y);
}
inline void cut(int x, int y) {
  access(y);
  splay(y);
  del(y, x);
  pushup(y);
}
void dfs(int x) {
  for (int i : g[x]) link(x, i), dfs(i);
}
int main() {
  n = read();
  for (int i = 2; i <= n; ++i) p[i] = read(), g[p[i]].push_back(i);
  for (int i = 1; i <= n; ++i)
    a[i] = read(), siz[i] = ssiz[i] = 1, sum[i] = a[i], ans[i] = a[i];
  dfs(1);
  splay(1);
  printf("%.10lf\n", 1.0 * ans[1] / n / n);
  m = read();
  while (m--) {
    char c;
    scanf(" %c", &c);
    int x = read(), y = read();
    if (c == 'P') {
      access(y);
      splay(y);
      splay(x);
      if (!gr(y)) swap(x, y);
      cut(y, p[y]);
      p[y] = x;
      link(x, y);
      splay(1);
    }
    if (c == 'V') access(x), splay(x), a[x] = y, pushup(x), splay(1);
    printf("%.10lf\n", 1.0 * ans[1] / n / n);
  }
  return 0;
}
