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 = 5e5;
struct Node {
  int ch[2], fa;
  int size, vsize;
  long long vans, svw, ans, vs2;
} a[N];
int fa[N], w[N], n, qn;
Node newnode(int w) {
  Node t;
  t.ans = w;
  t.size = 1;
  t.vsize = 1;
  t.vs2 = 0;
  t.svw = w;
  t.vans = 0;
  t.ch[0] = t.ch[1] = t.fa = 0;
  return t;
}
long long sqr(int x) { return 1ll * x * x; }
void pushup(int u) {
  int ls = a[u].ch[0], rs = a[u].ch[1];
  a[u].size = a[ls].size + a[rs].size + a[u].vsize;
  a[u].svw = a[ls].svw + a[rs].svw + 1ll * w[u] * a[u].vsize;
  a[u].ans = a[ls].ans + a[rs].ans + a[u].vans +
             1ll * w[u] *
                 (sqr(a[u].vsize) - a[u].vs2 + 2ll * a[rs].size * a[u].vsize) +
             2ll * a[ls].svw * (a[rs].size + a[u].vsize);
}
void upd(int u, int v, int f) {
  a[u].vsize += a[v].size * f;
  a[u].vs2 += sqr(a[v].size) * f;
  a[u].svw += 1ll * w[u] * a[v].size * f;
  a[u].vans += a[v].ans * f;
}
int getwh(int u) { return a[a[u].fa].ch[1] == u; }
int nroot(int x) { return a[a[x].fa].ch[0] == x || a[a[x].fa].ch[1] == x; }
void rotate(int u) {
  int v = a[u].fa, z = a[v].fa, wh = getwh(u), w = a[u].ch[wh ^ 1];
  if (nroot(v)) a[z].ch[getwh(v)] = u;
  a[v].ch[wh] = w, a[u].ch[wh ^ 1] = v;
  a[v].fa = u, a[u].fa = z, a[w].fa = v;
  pushup(v);
}
void splay(int u) {
  while (nroot(u)) {
    int v = a[u].fa;
    if (nroot(v)) {
      rotate(getwh(u) == getwh(v) ? v : u);
    }
    rotate(u);
  }
  pushup(u);
}
void access(int u) {
  for (int v = 0; u; v = u, u = a[u].fa)
    splay(u), upd(u, a[u].ch[1], 1), upd(u, v, -1), a[u].ch[1] = v, pushup(u);
}
void link(int u, int v) {
  access(v), splay(v), splay(u), a[u].fa = v;
  upd(v, u, 1);
  pushup(v);
}
void cut(int u, int v) {
  access(u);
  splay(v);
  a[v].ch[1] = 0;
  a[u].fa = 0;
  pushup(v);
}
int isanc(int x, int y) {
  access(y), splay(y), splay(x);
  return nroot(y);
}
void print(int u) {
  cout << u << " " << a[u].ch[0] << " " << a[u].ch[1] << " " << a[u].fa << " "
       << a[u].size << " " << a[u].vsize << " " << a[u].ans << endl;
  if (a[u].ch[0]) print(a[u].ch[0]);
  if (a[u].ch[1]) print(a[u].ch[1]);
}
void print() {
  for (int i = 1; i <= n; i++)
    if (!nroot(i)) print(i);
  puts("");
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) scanf("%d", fa + i);
  for (int i = 1; i <= n; i++) scanf("%d", w + i);
  for (int i = 1; i <= n; i++) a[i] = newnode(w[i]);
  for (int i = 2; i <= n; i++) link(i, fa[i]);
  splay(1);
  printf("%.10lf\n", (double)a[1].ans / (1ll * n * n));
  scanf("%d", &qn);
  for (int i = 1; i <= qn; i++) {
    char st[10];
    int x, y;
    scanf("%s", st + 1);
    scanf("%d%d", &x, &y);
    if (st[1] == 'P') {
      if (isanc(x, y)) swap(x, y);
      cut(x, fa[x]);
      fa[x] = y;
      link(x, y);
    } else {
      access(x), splay(x);
      w[x] = y;
      pushup(x);
    }
    splay(1);
    printf("%.10lf\n", (double)a[1].ans / (1ll * n * n));
  }
}
