Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 50005, L = 0, R = 1;
struct node;
extern node *nil;
struct node {
  long long v, size, size_sqr, lazy_size, sum, v1, v2;
  node *s[2], *f, *next, *rf;
  node()
      : v(0),
        size(0),
        size_sqr(0),
        lazy_size(0),
        sum(0),
        v1(0),
        v2(0),
        f(nil),
        next(nil),
        rf(nil) {
    s[L] = s[R] = nil;
  }
  bool root() { return f->s[L] != this && f->s[R] != this; }
  int get_dir() { return f->s[R] == this; }
  long long addsize(long long d) {
    if (this == nil) return 0;
    lazy_size += d;
    size += d;
    rf->size_sqr += d * (size + size - d);
    long long tmp = sum;
    sum += (v2 - v1) * d;
    return 2 * d * tmp;
  }
  long long changev(long long d) {
    node *tmp = s[R];
    push();
    for (tmp = s[R]; tmp->s[L] != nil; tmp = tmp->s[L]) tmp->push();
    long long delta = (d - v) * (size * size - size_sqr);
    sum += (d - v) * (size - tmp->size);
    long long tmpv = (d - v) * (-tmp->size);
    for (tmp = s[R]; tmp != nil; tmp = tmp->s[L]) tmp->sum += tmpv;
    v = d;
    update();
    return delta;
  }
  void push() {
    if (lazy_size) {
      s[L]->addsize(lazy_size);
      s[R]->addsize(lazy_size);
      lazy_size = 0;
    }
  }
  void update() {
    sum = s[L]->sum + s[R]->sum + size * (v - rf->v);
    v2 = s[R] != nil ? s[R]->v2 : v;
    v1 = s[L] != nil ? s[L]->v1 : rf->v;
  }
  void set_f(node *d) {
    if (this != nil) f = d;
  }
  void set_right(node *k) {
    push();
    node *tmp = s[R];
    for (; tmp != nil; tmp = tmp->s[L]) tmp->push();
    (s[R] = k)->set_f(this);
    for (tmp = s[R]; tmp->s[L] != nil; tmp = tmp->s[L])
      ;
    if (tmp != nil)
      for (; tmp != this; tmp = tmp->f) tmp->update();
    update();
  }
} u[N], *t[N] = {}, *nil = &u[0];
int n;
long long ans = 0;
vector<int> son[N];
int dfs(int s) {
  for (vector<int>::iterator it = son[s].begin(); it != son[s].end(); ++it) {
    long long tmp = dfs(*it);
    t[s]->size += tmp;
    t[s]->size_sqr += tmp * tmp;
  }
  t[s]->size += 1;
  ans += (t[s]->size * t[s]->size - t[s]->size_sqr) * t[s]->v;
  t[s]->sum += (t[s]->v - t[s]->rf->v) * t[s]->size;
  return t[s]->size;
}
void rotate(node *n) {
  node *f = n->f;
  int d = n->get_dir();
  f->push();
  n->push();
  f->s[d] = n->s[!d];
  f->s[d]->set_f(f);
  if (!f->root()) f->f->s[f->get_dir()] = n;
  n->f = f->f;
  n->s[!d] = f;
  f->f = n;
  f->update();
  n->update();
}
void Splay(node *n) {
  while (!n->root())
    if (n->f->root())
      rotate(n);
    else if (n->get_dir() ^ n->f->get_dir())
      rotate(n), rotate(n);
    else
      rotate(n->f), rotate(n);
}
node *Access(node *x) {
  node *y = nil;
  for (; x != nil; y = x, x = x->f) {
    Splay(x);
    x->set_right(y);
  }
  return y;
}
void init() {
  scanf("%d", &n);
  for (int i = 0; i <= n; ++i) t[i] = &u[i];
  for (int i = 2, x = 0; i <= n; ++i) {
    scanf("%d", &x);
    son[x].push_back(i);
    t[i]->f = t[i]->rf = t[x];
  }
  for (int i = 1; i <= n; ++i) scanf("%I64d", &t[i]->v);
}
void set_f(int x, int y) {
  node *p = t[x], *q = t[y], *r = p->rf;
  Splay(p);
  long long delta = p->size;
  ans += Access(r)->addsize(-delta) + r->v * delta * delta;
  r->size_sqr -= delta * delta;
  p->rf = p->f = q;
  ans += Access(q)->addsize(delta) + r->v * delta * delta;
  q->size_sqr += delta * delta;
}
void work() {
  int q, x, y;
  char ch;
  cin >> q;
  printf("%.9f\n", ans * 1.0 / n / n);
  for (int i = 1; i <= q; ++i) {
    cin >> ch >> x >> y;
    if (ch == 'P') {
      node *p = Access(t[y]), *q = t[x];
      while (!q->root()) q = q->f;
      p != q ? set_f(x, y) : set_f(y, x);
    }
    if (ch == 'V') {
      Access(t[x]);
      Splay(t[x]);
      ans += t[x]->changev(y);
    }
    printf("%.9f\n", ans * 1.0 / n / n);
  }
}
int main() {
  init();
  dfs(1);
  work();
  return 0;
}