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;
long long read() {
  long long x = 0, y = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') y = -y;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * y;
}
namespace qiqi {
const int N = 1e5 + 5;
int n, Q, ft[N], fa[N], ch[N][2];
long long a[N], sa[N], b[N], sb[N], ic[N], sc[N], sq[N];
inline char get() {
  char ch = getchar();
  while (ch != 'P' && ch != 'V') ch = getchar();
  return ch;
}
inline void push_up(int x) {
  sa[x] = a[x] * b[x] + sa[ch[x][0]] + sa[ch[x][1]];
  sb[x] = b[x] + sb[ch[x][0]] + sb[ch[x][1]];
  sc[x] = a[x] * (b[x] * b[x] - sq[x]) + ic[x] + sc[ch[x][0]] + sc[ch[x][1]] +
          2 * a[x] * b[x] * sb[ch[x][1]] +
          2 * sa[ch[x][0]] * (sb[x] - sb[ch[x][0]]);
}
inline bool isroot(int x) { return x != ch[fa[x]][0] && x != ch[fa[x]][1]; }
inline void rot(int x) {
  int y = fa[x], z = fa[y], d = x == ch[y][1];
  if (!isroot(y)) ch[z][y == ch[z][1]] = x;
  fa[x] = z;
  fa[y] = x;
  fa[ch[x][!d]] = y;
  ch[y][d] = ch[x][!d];
  ch[x][!d] = y;
  push_up(y);
  push_up(x);
}
inline void splay(int x) {
  int y;
  while (!isroot(x)) {
    if (!isroot(y = fa[x])) rot((x == ch[y][1]) != (y == ch[fa[y]][1]) ? x : y);
    rot(x);
  }
}
inline void access(int x) {
  for (int y = 0; x; x = fa[y = x]) {
    splay(x);
    b[x] += sb[ch[x][1]];
    ic[x] += sc[ch[x][1]];
    sq[x] += sb[ch[x][1]] * sb[ch[x][1]];
    ch[x][1] = y;
    b[x] -= sb[y];
    ic[x] -= sc[y];
    sq[x] -= sb[y] * sb[y];
    push_up(x);
  }
}
inline int lca(int x, int y) {
  access(x);
  access(y);
  splay(x);
  return fa[x] ? fa[x] : x;
}
inline void cut(int x) {
  access(x);
  splay(x);
  ch[x][0] = fa[ch[x][0]] = 0;
  push_up(x);
}
inline void link(int x) {
  access(ft[x]);
  splay(ft[x]);
  splay(x);
  fa[ch[x][0] = ft[x]] = x;
  push_up(x);
}
inline void upd(int x, int k) {
  access(x);
  splay(x);
  a[x] = k;
  push_up(x);
}
void main() {
  n = read();
  for (int i = (2); i <= (n); ++i) ft[i] = read();
  for (int i = (1); i <= (n); ++i) {
    b[i] = 1;
    a[i] = read();
    push_up(i);
  }
  for (int i = (2); i <= (n); ++i) link(i);
  access(1);
  splay(1);
  printf("%.10lf\n", 1.0 * sc[1] / n / n);
  Q = read();
  while (Q--) {
    char o = get();
    int x = read(), y = read();
    if (o == 'V')
      upd(x, y);
    else {
      if (lca(x, y) == x) swap(x, y);
      cut(x);
      ft[x] = y;
      link(x);
    }
    access(1);
    splay(1);
    printf("%.10lf\n", 1.0 * sc[1] / n / n);
  }
}
}  // namespace qiqi
int main() {
  qiqi::main();
  return 0;
}
