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;
int read();
const int N = 50005;
int n, m;
int FA[N];
int fa[N], ch[N][2];
long long a[N], sz[N], fsz[N], ans[N], saf[N], fans[N], sfs2[N];
int dir(int x) { return ch[fa[x]][1] == x; }
bool nrt(int x) { return ch[fa[x]][0] == x || ch[fa[x]][1] == x; }
void up(int x) {
  sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + fsz[x];
  saf[x] = saf[ch[x][0]] + saf[ch[x][1]] + a[x] * fsz[x];
  ans[x] = ans[ch[x][0]] + ans[ch[x][1]] + fans[x] +
           a[x] * (fsz[x] * fsz[x] - sfs2[x]) +
           2ll * a[x] * fsz[x] * sz[ch[x][1]] +
           2ll * saf[ch[x][0]] * (sz[x] - sz[ch[x][0]]);
}
void rot(int x) {
  int f = fa[x], g = fa[f], c = dir(x);
  if (nrt(f)) ch[g][dir(f)] = x;
  fa[x] = g;
  ch[f][c] = ch[x][c ^ 1];
  fa[ch[f][c]] = f;
  ch[x][c ^ 1] = f;
  fa[f] = x;
  up(f);
  up(x);
}
void splay(int x) {
  for (; nrt(x); rot(x))
    if (nrt(fa[x])) rot(dir(x) == dir(fa[x]) ? fa[x] : x);
}
void access(int x) {
  for (int y = 0; x; x = fa[y = x]) {
    splay(x);
    fsz[x] += sz[ch[x][1]];
    fans[x] += ans[ch[x][1]];
    sfs2[x] += sz[ch[x][1]] * sz[ch[x][1]];
    ch[x][1] = y;
    fsz[x] -= sz[ch[x][1]];
    fans[x] -= ans[ch[x][1]];
    sfs2[x] -= sz[ch[x][1]] * sz[ch[x][1]];
    up(x);
  }
}
void link(int x, int y) {
  access(x);
  splay(x);
  access(y);
  splay(y);
  fa[y] = x;
  fsz[x] += sz[y];
  fans[x] += ans[y];
  sfs2[x] += sz[y] * sz[y];
  up(x);
}
void cut(int x, int y) {
  access(x);
  splay(x);
  splay(y);
  fsz[x] -= sz[y];
  fans[x] -= ans[y];
  sfs2[x] -= sz[y] * sz[y];
  fa[y] = 0;
  up(x);
}
bool insubtree(int u, int v) {
  access(u);
  splay(u);
  splay(v);
  return nrt(u);
}
int main() {
  n = read();
  for (int(i) = (2); (i) <= (n); ++(i)) FA[i] = read();
  for (int(i) = (1); (i) <= (n); ++(i)) {
    a[i] = read();
    fsz[i] = sz[i] = 1;
    saf[i] = a[i] * fsz[i];
  }
  for (int(i) = (2); (i) <= (n); ++(i)) link(FA[i], i);
  access(1);
  splay(1);
  printf("%.9lf\n", 1. * ans[1] / n / n);
  m = read();
  for (int i = 1, x, y; i <= m; ++i) {
    char c;
    scanf(" %c", &c);
    x = read(), y = read();
    if (c == 'P') {
      if (insubtree(y, x)) swap(x, y);
      cut(FA[x], x);
      link(y, x);
      FA[x] = y;
      access(1);
      splay(1);
      printf("%.9lf\n", 1. * ans[1] / n / n);
    } else {
      access(x);
      splay(x);
      a[x] = y;
      up(x);
      printf("%.9lf\n", 1. * ans[x] / n / n);
    }
  }
  return 0;
}
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return f * x;
}
