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 = 2e6 + 5;
struct node {
  int child[2], fa;
  long long sum, ans, ad, de, all, size;
} tree[N];
int n, m, a[N], fa[N];
inline void upd(int p) {
  tree[p].sum =
      tree[tree[p].child[0]].sum + tree[tree[p].child[1]].sum + tree[p].size;
  tree[p].ans = tree[tree[p].child[0]].ans + tree[tree[p].child[1]].ans +
                tree[p].ad +
                (1LL * tree[p].size * tree[p].size - tree[p].de) * a[p] +
                2LL * tree[p].size * tree[tree[p].child[1]].sum * a[p] +
                2LL * tree[tree[p].child[0]].all *
                    (tree[p].sum - tree[tree[p].child[0]].sum);
  tree[p].all = tree[tree[p].child[0]].all + tree[tree[p].child[1]].all +
                tree[p].size * a[p];
}
inline int son_check(int p) { return tree[tree[p].fa].child[1] == p; }
inline bool root_check(int p) {
  return tree[tree[p].fa].child[1] != p && tree[tree[p].fa].child[0] != p;
}
void rotate(int p) {
  int f = tree[p].fa, g = tree[f].fa, k = son_check(p);
  !root_check(f) && (tree[g].child[son_check(f)] = p), tree[p].fa = g;
  tree[f].child[k] = tree[p].child[k ^ 1], tree[tree[p].child[k ^ 1]].fa = f;
  tree[p].child[k ^ 1] = f, tree[f].fa = p;
  upd(f), upd(p);
}
void splay(int p) {
  int u = p;
  while (!root_check(p)) {
    int f = tree[p].fa;
    if (!root_check(f)) rotate(son_check(p) == son_check(f) ? f : p);
    rotate(p);
  }
}
inline void smodify(int p, int s, int x) {
  tree[p].size += x * tree[s].sum;
  tree[p].de += x * tree[s].sum * tree[s].sum;
  tree[p].ad += x * tree[s].ans;
}
inline void access(int p) {
  for (int x = 0; p; p = tree[x = p].fa)
    splay(p), smodify(p, tree[p].child[1], 1),
        tree[p].child[1] = x, smodify(p, tree[p].child[1], -1), upd(p);
}
inline void link(int x, int y) {
  access(y);
  splay(y);
  access(x);
  splay(x);
  tree[y].fa = x;
  smodify(x, y, 1);
  upd(x);
}
inline void cut(int x, int y) {
  access(x);
  splay(x);
  splay(y);
  tree[y].fa = 0;
  smodify(x, y, -1);
  upd(x);
}
inline bool ancesstor(int x, int y) {
  access(y);
  splay(y);
  splay(x);
  if (root_check(y)) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) scanf("%d", &fa[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    tree[i].all = tree[i].ans = a[i];
    tree[i].sum = tree[i].size = 1;
  }
  for (int i = 2; i <= n; i++) link(fa[i], i);
  access(1);
  splay(1);
  int _;
  double __ = 1LL * n * n, Ans = tree[1].ans / __;
  printf("%0.8lf\n", Ans);
  scanf("%d", &_);
  while (_--) {
    int opt, x, y;
    char _[10];
    scanf("%s", _ + 1);
    if (_[1] == 'P')
      opt = 1;
    else
      opt = 2;
    scanf("%d%d", &x, &y);
    if (opt == 1) {
      if (!ancesstor(x, y)) {
        cut(fa[x], x);
        link(y, x);
        fa[x] = y;
      } else {
        cut(fa[y], y);
        link(x, y);
        fa[y] = x;
      }
      access(1);
      splay(1);
      Ans = tree[1].ans / __;
    } else {
      access(x);
      splay(x);
      a[x] = y;
      upd(x);
      Ans = tree[x].ans / __;
    }
    printf("%0.8lf\n", Ans);
  }
  return 0;
}