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, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f = ch == '-', ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
const int N = 50005;
int n, q;
int p[N];
int fa[N], son[N][2];
long long a[N], vsz[N], vsz2[N], vans[N], ans[N], sz[N], csum[N];
int isroot(int x) { return son[fa[x]][0] != x && son[fa[x]][1] != x; }
int wson(int x) { return son[fa[x]][1] == x; }
void pushup(int x) {
  int ls = son[x][0], rs = son[x][1];
  sz[x] = vsz[x] + sz[ls] + sz[rs];
  csum[x] = vsz[x] * a[x] + csum[ls] + csum[rs];
  ans[x] = ans[ls] + ans[rs] + vans[x] + (vsz[x] * vsz[x] - vsz2[x]) * a[x] +
           2LL * vsz[x] * sz[rs] * a[x] + 2LL * csum[ls] * (sz[rs] + vsz[x]);
}
void rotate(int x) {
  int y = fa[x], z = fa[y], L = wson(x), R = L ^ 1;
  assert(!isroot(x));
  if (!isroot(y)) son[z][wson(y)] = x;
  fa[x] = z, fa[y] = x, fa[son[x][R]] = y;
  son[y][L] = son[x][R], son[x][R] = y;
  pushup(y), pushup(x);
}
void splay(int x) {
  for (int y = fa[x]; !isroot(x); rotate(x), y = fa[x])
    if (!isroot(y)) rotate(wson(x) == wson(y) ? y : x);
}
void access(int x) {
  int t = 0;
  while (x) {
    splay(x);
    vsz[x] += sz[son[x][1]];
    vsz2[x] += sz[son[x][1]] * sz[son[x][1]];
    vans[x] += ans[son[x][1]];
    son[x][1] = t;
    vsz[x] -= sz[son[x][1]];
    vsz2[x] -= sz[son[x][1]] * sz[son[x][1]];
    vans[x] -= ans[son[x][1]];
    pushup(x), t = x, x = fa[x];
  }
}
void setfather(int x, int f) {
  access(x), access(f), splay(f);
  vsz[f] += sz[x];
  vsz2[f] += sz[x] * sz[x];
  vans[f] += ans[x];
  fa[x] = p[x] = f;
  pushup(f);
}
void cut(int x) {
  int f = p[x];
  access(x), access(f), splay(x);
  assert(fa[x] == f && isroot(x));
  vsz[f] -= sz[x];
  vsz2[f] -= sz[x] * sz[x];
  vans[f] -= ans[x];
  fa[x] = p[x] = 0;
  pushup(f);
}
int insubtree(int x, int f) {
  access(x), splay(x);
  int y = f;
  while (!isroot(y)) y = fa[y];
  splay(f);
  return y == x;
}
int main() {
  n = read();
  for (int i = (2); i <= (n); i++) p[i] = read();
  for (int i = (1); i <= (n); i++) a[i] = read();
  for (int i = (1); i <= (n); i++)
    vsz[i] = sz[i] = 1, vsz2[i] = vans[i] = 0, ans[i] = csum[i] = a[i];
  for (int i = (2); i <= (n); i++) setfather(i, p[i]);
  q = read();
  splay(1);
  printf("%.10lf\n", (double)ans[1] / n / n);
  while (q--) {
    char s[5];
    scanf("%s", s);
    if (s[0] == 'P') {
      int x = read(), f = read();
      if (insubtree(f, x)) swap(x, f);
      cut(x);
      setfather(x, f);
    } else if (s[0] == 'V') {
      int x = read(), v = read();
      access(x), splay(x);
      a[x] = v;
      pushup(x);
    }
    splay(1);
    printf("%.10lf\n", (double)ans[1] / n / n);
  }
  return 0;
}
