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>
const int MAXN = 1e5 + 10;
inline long long pw(int x) { return (long long)x * x; }
int son[MAXN][2], fa[MAXN];
int get(int x, int b = 1) { return son[fa[x]][b] == x; }
int isroot(int x) { return !(get(x, 0) || get(x, 1)); }
int allv[MAXN];
struct info {
  long long a, b;
  int sz;
  friend info operator+(const info& lhs, const info& rhs) {
    info res;
    res.a = lhs.a + rhs.a;
    res.b = lhs.b + rhs.b + lhs.a * rhs.sz;
    res.sz = lhs.sz + rhs.sz;
    return res;
  }
} val[MAXN], vt[MAXN];
void add(info& x, const info& y, int co) {
  x.b += y.b * co, x.sz += y.sz * co;
  x.a += pw(y.sz) * co;
}
void update(int x) {
  val[x].sz = vt[x].sz + 1;
  val[x].a = val[x].sz * 2ll * allv[x];
  val[x].b = vt[x].b + (pw(val[x].sz) - vt[x].a) * allv[x];
  if (int t = son[x][0]) val[x] = val[t] + val[x];
  if (int t = son[x][1]) val[x] = val[x] + val[t];
}
void rotate(int x) {
  int y = fa[x], z = fa[y];
  bool b = get(x);
  if (!isroot(y)) son[z][get(y)] = x;
  son[y][b] = son[x][!b], son[x][!b] = y;
  fa[x] = z, fa[y] = x;
  if (int t = son[y][b]) fa[t] = y;
  update(y);
}
void splay(int x) {
  for (; !isroot(x); rotate(x))
    if (!isroot(fa[x])) rotate(get(x) ^ get(fa[x]) ? x : fa[x]);
  update(x);
}
void access(int x) {
  int t = 0;
  while (x) {
    splay(x);
    if (int v = son[x][1]) add(vt[x], val[v], 1);
    son[x][1] = t;
    add(vt[x], val[t], -1);
    update(x);
    t = x, x = fa[x];
  }
}
void link(int x, int y) {
  access(y);
  splay(y);
  splay(x);
  fa[x] = y;
  add(vt[y], val[x], 1);
  update(y);
}
void cut(int x) {
  access(x);
  splay(x);
  int y = son[x][0];
  fa[y] = son[x][0] = 0;
  update(x);
}
bool conn(int x, int y) {
  splay(y);
  while (son[y][0]) y = son[y][0];
  splay(y);
  return x == y;
}
int n;
void printans() {
  access(1);
  std::cout << (val[1].b / (double)n / n) << '\n';
}
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  std::cout << std::fixed << std::setprecision(10);
  std::cin >> n;
  static int tf[MAXN];
  for (int i = 2; i <= n; ++i) std::cin >> tf[i];
  for (int i = 1; i <= n; ++i) std::cin >> allv[i], update(i);
  for (int i = 2; i <= n; ++i) link(i, tf[i]);
  printans();
  int Q;
  std::cin >> Q;
  while (Q-- > 0) {
    static char opt[10];
    int x, y;
    std::cin >> opt >> x >> y;
    if (*opt == 'V') {
      access(x);
      splay(x);
      allv[x] = y;
      update(x);
    } else {
      access(y);
      access(tf[x]);
      if (conn(x, y)) std::swap(x, y);
      cut(x);
      link(x, y);
      tf[x] = y;
    }
    printans();
  }
  return 0;
}
