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 N = 50005;
int n, m;
int f[N], ch[N][2], val[N];
long long ans[N], vans[N], vsum2[N], all[N];
int sum[N], vsum[N];
long long sqr(int x) { return (long long)x * x; }
bool get(int x) { return ch[f[x]][1] == x; }
bool isroot(int x) { return ch[f[x]][0] != x && ch[f[x]][1] != x; }
void pushup(int x) {
  int L = ch[x][0], R = ch[x][1];
  sum[x] = sum[L] + sum[R] + vsum[x];
  all[x] = all[L] + all[R] + (long long)val[x] * vsum[x];
  ans[x] = ans[L] + ans[R] + vans[x] + 2LL * all[L] * (sum[x] - sum[L]) +
           2LL * val[x] * vsum[x] * sum[R] + val[x] * (sqr(vsum[x]) - vsum2[x]);
}
void rotate(int x) {
  int y = f[x], z = f[y];
  bool b = get(x);
  if (!isroot(y)) ch[z][get(y)] = x;
  ch[y][b] = ch[x][!b], ch[x][!b] = y;
  f[x] = z, f[y] = x;
  if (ch[y][b]) f[ch[y][b]] = y;
  pushup(y);
}
void splay(int x) {
  for (; !isroot(x); rotate(x))
    if (!isroot(f[x])) rotate(get(x) ^ get(f[x]) ? x : f[x]);
  pushup(x);
}
void update(int x, int y, int v = 1) {
  vsum[x] += v * sum[y];
  vsum2[x] += (long long)v * sum[y] * sum[y];
  vans[x] += v * ans[y];
}
void access(int x) {
  for (int t = 0; x; t = x, x = f[x]) {
    splay(x);
    update(x, t, -1);
    update(x, ch[x][1]);
    ch[x][1] = t, pushup(x);
  }
}
void makeroot(int x) { access(x), splay(x); }
void link(int x, int y) {
  makeroot(x), makeroot(y);
  f[x] = y, update(y, x);
  pushup(y);
}
void cut(int x) {
  makeroot(x);
  f[ch[x][0]] = 0, ch[x][0] = 0;
  pushup(x);
}
int find_root(int x) {
  splay(x);
  while (ch[x][0]) x = ch[x][0];
  return splay(x), x;
}
bool ancestor(int x, int y) {
  access(y);
  return find_root(x) == find_root(y);
}
void modify(int x, int v) {
  makeroot(x);
  val[x] = v, pushup(x);
}
void query() {
  access(1);
  std::cout << std::fixed << std::setprecision(10) << (double)ans[1] / n / n
            << '\n';
}
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0);
  std::cin >> n;
  for (int i = 1; i <= n; ++i) vsum[i] = 1, pushup(i);
  for (int i = 2, x; i <= n; ++i) std::cin >> x, link(i, x);
  for (int i = 1, x; i <= n; ++i) std::cin >> x, modify(i, x);
  std::cin >> m, query();
  for (int i = 1; i <= m; ++i) {
    char opt;
    int x, y;
    std::cin >> opt >> x >> y;
    if (opt == 'P') {
      if (ancestor(x, y))
        cut(y), link(y, x);
      else
        cut(x), link(x, y);
    } else
      modify(x, y);
    query();
  }
  return 0;
}
