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 MAXN = 50 * 1000 + 5, K = 500;
int p[MAXN], comp[MAXN], snd[MAXN];
char root[MAXN];
long long s[MAXN], ts[MAXN], pv[MAXN], pvs[MAXN], sv[MAXN];
pair<char, pair<int, int> > q[MAXN];
vector<int> g[MAXN], gc[MAXN];
bool dfsLca(int v) {
  int cnt = root[v];
  for (size_t i = 0; i < g[v].size(); i++) cnt += dfsLca(g[v][i]);
  if (cnt > 1) root[v] = true;
  return cnt;
}
void dfs0(int v) {
  ts[v] = 1;
  sv[v] = 1;
  for (size_t i = 0; i < g[v].size(); i++) {
    dfs0(g[v][i]);
    sv[v] += 2ll * ts[g[v][i]] * ts[v];
    ts[v] += ts[g[v][i]];
  }
}
void dfs1(int v, int c) {
  comp[v] = c;
  if (root[v])
    pvs[v] = 0;
  else {
    pvs[v] = pv[v] + pvs[p[v]];
    snd[v] = (root[p[v]] ? v : snd[p[v]]);
  }
  for (size_t i = 0; i < g[v].size(); i++)
    if (!root[g[v][i]]) dfs1(g[v][i], c);
}
bool anc(int v, int u) {
  for (int w = u;; w = comp[p[w]]) {
    if (w == v) return true;
    if (w == 0) return false;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(12);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> p[i];
    p[i]--;
    g[p[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> s[i];
  dfs0(0);
  for (int i = 1; i < n; i++) pv[i] = (ts[p[i]] - ts[i]) * s[p[i]];
  long long ans = 0;
  for (int i = 0; i < n; i++) ans += s[i] * sv[i];
  cout << (long double)ans / ((long long)n * n) << '\n';
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> q[i].first >> q[i].second.first >> q[i].second.second;
    q[i].second.first--;
    if (q[i].first == 'P') q[i].second.second--;
  }
  for (int bl = 0; bl < m; bl += K) {
    int br = min(m, bl + K);
    for (int i = 0; i < n; i++) root[i] = false;
    root[0] = true;
    for (int i = bl; i < br; i++) {
      root[q[i].second.first] = true;
      if (q[i].first == 'P') root[q[i].second.second] = true;
    }
    dfsLca(0);
    dfs0(0);
    for (int i = 1; i < n; i++) pv[i] = 2 * (ts[p[i]] - ts[i]) * s[p[i]];
    for (int i = 0; i < n; i++)
      if (root[i]) dfs1(i, i);
    vector<int> roots;
    for (int i = 0; i < n; i++)
      if (root[i]) roots.push_back(i);
    int k = roots.size();
    for (int i = 0; i < k; i++)
      if (roots[i] != 0) gc[comp[p[roots[i]]]].push_back(roots[i]);
    for (int i = bl; i < br; i++) {
      if (q[i].first == 'P') {
        int v = q[i].second.first, u = q[i].second.second;
        if (anc(v, u)) swap(v, u);
        for (int w = v; w != 0; w = comp[p[w]]) {
          ans -= ts[v] * (pv[w] + pvs[p[w]]);
          if (p[w] != comp[p[w]]) ts[snd[p[w]]] -= ts[v];
          ts[comp[p[w]]] -= ts[v];
          sv[comp[p[w]]] -=
              2 *
              (ts[comp[p[w]]] -
               (p[w] == comp[p[w]] ? (w == v ? 0 : ts[w]) : ts[snd[p[w]]])) *
              ts[v];
          for (size_t j = 0; j < gc[comp[p[w]]].size(); j++)
            if (gc[comp[p[w]]][j] != w) {
              if (p[gc[comp[p[w]]][j]] == comp[p[w]])
                pv[gc[comp[p[w]]][j]] -= 2 * ts[v] * s[comp[p[w]]];
              else
                pvs[p[gc[comp[p[w]]][j]]] -= 2 * ts[v] * s[comp[p[w]]];
            }
        }
        for (size_t j = 0; j < gc[comp[p[v]]].size(); j++)
          if (gc[comp[p[v]]][j] == v) {
            gc[comp[p[v]]].erase(gc[comp[p[v]]].begin() + j);
            break;
          }
        p[v] = u;
        gc[u].push_back(v);
        pv[v] = 2 * ts[u] * s[u];
        for (int w = v; w != 0; w = comp[p[w]]) {
          ans += ts[v] * (pv[w] + pvs[p[w]]);
          if (p[w] != comp[p[w]]) ts[snd[p[w]]] += ts[v];
          ts[comp[p[w]]] += ts[v];
          sv[comp[p[w]]] +=
              2 *
              (ts[comp[p[w]]] - (p[w] == comp[p[w]] ? ts[w] : ts[snd[p[w]]])) *
              ts[v];
          for (size_t j = 0; j < gc[comp[p[w]]].size(); j++)
            if (gc[comp[p[w]]][j] != w) {
              if (p[gc[comp[p[w]]][j]] == comp[p[w]])
                pv[gc[comp[p[w]]][j]] += 2 * ts[v] * s[comp[p[w]]];
              else
                pvs[p[gc[comp[p[w]]][j]]] += 2 * ts[v] * s[comp[p[w]]];
            }
        }
      } else {
        int v = q[i].second.first, x = q[i].second.second;
        ans += (x - s[v]) * sv[v];
        for (size_t j = 0; j < gc[v].size(); j++)
          if (p[gc[v][j]] == v)
            pv[gc[v][j]] += 2 * (x - s[v]) * (ts[v] - ts[gc[v][j]]);
          else
            pvs[p[gc[v][j]]] += 2 * (x - s[v]) * (ts[v] - ts[snd[p[gc[v][j]]]]);
        s[v] = x;
      }
      cout << (long double)ans / ((long long)n * n) << '\n';
    }
    for (int i = 0; i < n; i++) {
      g[i].clear();
      gc[i].clear();
    }
    for (int i = 1; i < n; i++) g[p[i]].push_back(i);
  }
  return 0;
}
