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 N = 50005;
int n, fa[N], p[N], ch[N][2], val[N], sz[N], dsz[N], tag[N];
long long w[N], sum[N], ans;
vector<int> g[N];
void upd(int k1) {
  sz[k1] = dsz[k1] + sz[ch[k1][0]] + sz[ch[k1][1]];
  sum[k1] = 1ll * val[k1] * dsz[k1] + sum[ch[k1][0]] + sum[ch[k1][1]];
}
void pst(int k1, int k2) { tag[k1] += k2, w[k1] += 1ll * dsz[k1] * k2; }
void psd(int k1) {
  if (tag[k1]) pst(ch[k1][0], tag[k1]), pst(ch[k1][1], tag[k1]), tag[k1] = 0;
}
bool isr(int k1) { return ch[fa[k1]][0] != k1 && ch[fa[k1]][1] != k1; }
int get(int k1) { return ch[fa[k1]][1] == k1; }
void rtt(int k1) {
  int k2 = fa[k1], k3 = get(k1);
  if (!isr(k2)) ch[fa[k2]][get(k2)] = k1;
  fa[k1] = fa[k2];
  ch[k2][k3] = ch[k1][k3 ^ 1], fa[ch[k1][k3 ^ 1]] = k2;
  ch[k1][k3 ^ 1] = k2, fa[k2] = k1;
  upd(k2), upd(k1);
}
void splay(int k1) {
  static int q[N];
  q[*q = 1] = k1;
  for (int i = k1; !isr(i); q[++*q] = fa[i], i = fa[i])
    ;
  for (int i = *q; i >= 1; --i) psd(q[i]);
  for (; !isr(k1); rtt(k1))
    if (!isr(fa[k1])) rtt(get(k1) == get(fa[k1]) ? fa[k1] : k1);
}
void access(int k1) {
  for (int i = 0; k1; i = k1, k1 = fa[k1]) {
    splay(k1), dsz[k1] += sz[ch[k1][1]];
    ch[k1][1] = i, dsz[k1] -= sz[ch[k1][1]];
    upd(k1);
  }
}
void dfs(int k1) {
  w[k1] = dsz[k1] = 1;
  for (int j : g[k1])
    dfs(j), w[k1] += 2ll * dsz[k1] * dsz[j], dsz[k1] += dsz[j];
  ans += w[k1] * val[k1];
  upd(k1);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i)
    scanf("%d", &fa[i]), g[fa[i]].push_back(i), p[i] = fa[i];
  for (int i = 1; i <= n; ++i) scanf("%d", &val[i]);
  dfs(1);
  printf("%.20lf\n", 1.0 * ans / n / n);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    char c;
    while (!isupper(c = getchar()))
      ;
    int k1, k2;
    scanf("%d%d", &k1, &k2);
    if (c == 'P') {
      access(k2), splay(k1);
      if (!fa[k1]) swap(k1, k2);
      int k3 = p[k1];
      access(k3), splay(k3);
      dsz[k3] -= sz[k1];
      pst(k3, -sz[k1] * 2);
      upd(k3), ans -= sum[k3] * sz[k1] * 2;
      access(k2), splay(k2);
      ans += sum[k2] * sz[k1] * 2;
      pst(k2, sz[k1] * 2);
      dsz[k2] += sz[k1];
      upd(k2);
      fa[k1] = k2;
      p[k1] = k2;
    } else {
      splay(k1);
      ans += (k2 - val[k1]) * w[k1];
      val[k1] = k2;
    }
    printf("%.20lf\n", 1.0 * ans / n / n);
  }
  return 0;
}
