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 S = 55555;
int c[2][S], f[S], n, m, tf[S];
long long w[S], iw[S], as[S], a[S], iss[S], is[S], s[S];
char o[3];
void up(int p) {
  s[p] = is[p] + s[c[0][p]] + s[c[1][p]],
  as[p] = as[c[0][p]] + as[c[1][p]] + a[p] * is[p],
  w[p] = w[c[0][p]] + w[c[1][p]] + iw[p] + a[p] * (is[p] * is[p] - iss[p]) +
         2 * a[p] * s[c[1][p]] * is[p] + 2 * as[c[0][p]] * (s[p] - s[c[0][p]]);
}
bool nr(int p) { return c[0][f[p]] == p || c[1][f[p]] == p; }
void spin(int p) {
  int F = f[p], G = f[F], d = c[1][F] == p, B = c[!d][p];
  if (nr(F)) c[c[1][G] == F][G] = p;
  c[d][F] = B;
  c[!d][p] = F;
  f[f[f[B] = F] = p] = G;
  up(F);
  up(p);
}
void ctb(int p, int r = 1) {
  iw[f[p]] += r * w[p];
  iss[f[p]] += r * s[p] * s[p];
  is[f[p]] += r * s[p];
}
void splay(int p) {
  for (; nr(p); spin(p))
    ;
  up(p);
}
void access(int r) {
  for (int y = 0, p = r; p; p = f[y = p])
    splay(p), ctb(c[1][p]), ctb(y, -1), c[1][p] = y, up(p);
  splay(r);
}
void link(int a, int b) {
  access(b);
  access(a);
  f[b] = a;
  ctb(b);
  up(a);
}
void cut(int p) {
  access(p);
  f[c[0][p]] = 0;
  c[0][p] = 0;
  up(p);
}
bool chk(int a, int b) {
  access(a);
  splay(b);
  return nr(a);
}
void Ans() {
  access(1);
  printf("%.9lf\n", (double)w[1] / n / n);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) scanf("%d", &tf[i]);
  for (int i = 1; i <= n; ++i)
    scanf("%lld", &a[i]), s[i] = is[i] = 1, w[i] = as[i] = a[i];
  for (int i = 2; i <= n; ++i) link(tf[i], i);
  Ans();
  scanf("%d", &m);
  for (int _ = 1, x, y; _ <= m; ++_, Ans()) {
    scanf("%s%d%d", o, &x, &y);
    if (o[0] == 'V')
      access(x), a[x] = y, up(x);
    else if (chk(y, x))
      cut(y), link(x, y);
    else
      cut(x), link(y, x);
  }
}