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>
using namespace std;
const int INF = (1 << 30) - 1;
const long long LINF = (1LL << 61) - 1;
const double EPS = 1e-10;
const int N = 5e4 + 100;
int ch[N][2], fa[N];
long long siz[N], couple[N], sum[N], path[N], lazy[N], val[N];
void pushup(int u) {
  sum[u] = sum[ch[u][0]] + sum[ch[u][1]] + val[u] * path[u];
}
void add(int u, long long c) {
  if (!u) return;
  siz[u] += c;
  couple[u] += 2 * path[u] * c;
  lazy[u] += c;
}
void pushdown(int u) {
  if (!lazy[u]) return;
  add(ch[u][0], lazy[u]);
  add(ch[u][1], lazy[u]);
  lazy[u] = 0;
}
void pushpath(int x) {
  if (!(ch[fa[x]][0] != x && ch[fa[x]][1] != x)) pushpath(fa[x]);
  pushdown(x);
}
void rot(int x) {
  int y = fa[x], z = fa[y];
  int k = ch[y][1] == x;
  if (!(ch[fa[y]][0] != y && ch[fa[y]][1] != y)) ch[z][ch[z][1] == y] = x;
  ch[y][k] = ch[x][k ^ 1];
  ch[x][k ^ 1] = y;
  fa[ch[y][k]] = y;
  fa[y] = x;
  fa[x] = z;
  pushup(y);
  pushup(x);
}
void splay(int x) {
  pushpath(x);
  while (!(ch[fa[x]][0] != x && ch[fa[x]][1] != x)) {
    int y = fa[x], z = fa[y];
    if ((ch[fa[y]][0] != y && ch[fa[y]][1] != y)) {
      rot(x);
    } else {
      rot(ch[y][0] == x ^ ch[z][0] == y ? x : y), rot(x);
    }
  }
}
int access(int x) {
  int last = 0;
  while (x) {
    if (last) {
      while (ch[last][0]) last = ch[last][0];
      splay(last);
    }
    splay(x);
    ch[x][1] = last;
    fa[last] = x;
    path[x] = siz[x] - siz[last];
    pushup(x);
    last = x;
    x = fa[x];
  }
  return last;
}
int n, q;
vector<int> son[N];
long long ans;
void dfs(int u) {
  couple[u] = 1, siz[u] = 1;
  for (int i = 0; i < son[u].size(); i++) {
    int v = son[u][i];
    dfs(v);
    couple[u] += 2 * siz[v] * siz[u];
    siz[u] += siz[v];
  }
  path[u] = siz[u];
  pushup(u);
  ans += couple[u] * val[u];
}
void dfsout(int u) {
  if (!u) return;
  dfsout(ch[u][0]);
  printf("%d ", u);
  dfsout(ch[u][1]);
}
int findroot(int u) {
  access(u);
  while (ch[u][0]) u = ch[u][0];
  return u;
}
int main() {
  srand((int)time(0));
  ios::sync_with_stdio(false);
  cin.tie(0);
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", fa + i);
    son[fa[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) scanf("%lld", &val[i]);
  dfs(1);
  printf("%.15lf\n", ans * 1.0 / n / n);
  scanf("%d", &q);
  char tp[5];
  int u, v;
  while (q--) {
    scanf("%s%d%d", tp, &u, &v);
    if (tp[0] == 'P') {
      access(v);
      splay(u);
      if (!fa[u]) swap(u, v);
      access(u);
      splay(u);
      int ls = ch[u][0];
      ans -= 2 * sum[ls] * siz[u];
      add(ls, -siz[u]);
      fa[ls] = 0;
      ch[u][0] = 0;
      pushup(u), pushup(ls);
      access(v);
      splay(v);
      ans += 2 * sum[v] * siz[u];
      add(v, siz[u]);
      fa[u] = v;
      pushup(v);
    } else {
      access(u);
      ans += couple[u] * (v - val[u]);
      val[u] = v;
      pushup(u);
    }
    printf("%.15lf\n", ans * 1.0 / n / n);
  }
  return 0;
}
