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;
int n, m, tp;
int fa[N], ch[N][2];
;
long long ans;
struct node {
  int dsz, sz, tg, v;
  long long w, sum;
} p[N];
void pushup(int k) {
  int l = ch[k][0], r = ch[k][1];
  p[k].sz = p[k].dsz + p[l].sz + p[r].sz;
  p[k].sum = 1ll * p[k].dsz * p[k].v + p[l].sum + p[r].sum;
}
void change(int k, int y) {
  p[k].tg += y;
  p[k].w += 1ll * y * p[k].dsz;
}
void pushdown(int k) {
  if (!p[k].tg) return;
  change(ch[k][0], p[k].tg);
  change(ch[k][1], p[k].tg);
  p[k].tg = 0;
}
bool isrt(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
void rotate(int x) {
  int y = fa[x], z = fa[y], l = (ch[y][1] == x), r = l ^ 1;
  if (!isrt(y)) ch[z][ch[z][1] == y] = x;
  fa[x] = z;
  fa[y] = x;
  fa[ch[x][r]] = y;
  ch[y][l] = ch[x][r];
  ch[x][r] = y;
  pushup(y);
}
void splay(int x) {
  static int q[N], top = 0;
  q[top = 1] = x;
  for (int i = x; !isrt(i); q[++top] = i = fa[i])
    ;
  for (; top; pushdown(q[top--]))
    ;
  for (; !isrt(x); rotate(x)) {
    int y = fa[x], z = fa[y];
    if (!isrt(y)) {
      if ((ch[z][0] == y) ^ (ch[y][0] == x))
        rotate(x);
      else
        rotate(y);
    }
  }
  pushup(x);
}
void access(int x) {
  int y = 0;
  for (; x; y = x, x = fa[x]) {
    splay(x);
    p[x].dsz += p[ch[x][1]].sz;
    ch[x][1] = y;
    p[x].dsz -= p[y].sz;
    pushup(x);
  }
}
struct edge {
  int to, next;
} e[N * 2];
int head[N], tot;
void add(int x, int y) {
  e[++tot] = (edge){y, head[x]};
  head[x] = tot;
}
void dfs(int x) {
  node &o = p[x];
  o.w = o.dsz = 1;
  for (int i = head[x]; i; i = e[i].next) {
    int y = e[i].to;
    dfs(e[i].to);
    o.w += 1ll * o.dsz * p[y].dsz * 2;
    o.dsz += p[y].dsz;
  }
  ans += o.v * o.w;
  pushup(x);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &fa[i]);
    add(fa[i], i);
  }
  for (int i = 1; i <= n; i++) scanf("%d", &p[i].v);
  dfs(1);
  printf("%.20lf\n", 1.0 * ans / n / n);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    char s[5];
    int x, y, z;
    scanf("%s%d%d", s, &x, &y);
    if (s[0] == 'P') {
      access(y);
      splay(x);
      for (z = x; !isrt(z); z = ch[z][0])
        ;
      if (fa[z])
        access(x);
      else
        swap(x, y);
      if (!ch[x][0])
        z = fa[x];
      else
        for (z = ch[x][0]; ch[z][1]; z = ch[z][1])
          ;
      if (z != y) {
        access(z);
        splay(z);
        splay(x);
        p[z].dsz -= p[x].dsz;
        change(z, -(p[x].sz * 2));
        pushup(z);
        ans -= p[z].sum * p[x].sz * 2;
        access(y);
        splay(y);
        ans += p[y].sum * p[x].sz * 2;
        change(y, p[x].sz * 2);
        p[y].dsz += p[x].sz;
        pushup(y);
        fa[x] = y;
      }
    } else {
      splay(x);
      ans += (y - p[x].v) * p[x].w;
      p[x].v = y;
    }
    printf("%.20lf\n", 1.0 * ans / n / n);
  }
}