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 arriz = 1.1e6;
int n, m, pre[arriz];
long long Ans;
struct node {
  int child[2], fa, siz, sum;
  long long ans, all, ad, de, a;
} tr[arriz];
void pushup(int o) {
  tr[o].sum = tr[tr[o].child[0]].sum + tr[tr[o].child[1]].sum + tr[o].siz;
  tr[o].all =
      tr[tr[o].child[0]].all + tr[tr[o].child[1]].all + tr[o].a * tr[o].siz;
  tr[o].ans = tr[tr[o].child[0]].ans + tr[tr[o].child[1]].ans + tr[o].ad;
  tr[o].ans += tr[o].a * (1ll * tr[o].siz * tr[o].siz - tr[o].de);
  tr[o].ans += 2ll * tr[o].a * tr[o].siz * tr[tr[o].child[1]].sum;
  tr[o].ans +=
      2ll * tr[tr[o].child[0]].all * (tr[o].sum - tr[tr[o].child[0]].sum);
}
bool id(int o) { return tr[tr[o].fa].child[1] == o; }
bool nr(int o) {
  return tr[tr[o].fa].child[1] == o || tr[tr[o].fa].child[0] == o;
}
void rotate(int o) {
  int x = tr[o].fa, y = tr[x].fa, fg = id(o), b = tr[o].child[!fg];
  if (nr(x)) tr[y].child[id(x)] = o;
  tr[o].fa = y;
  tr[x].child[fg] = b, tr[b].fa = x, tr[o].child[!fg] = x, tr[x].fa = o;
  return pushup(x), pushup(o);
}
void splay(int o) {
  for (; nr(o); rotate(o))
    if (nr(tr[o].fa)) id(o) ^ id(tr[o].fa) ? rotate(o) : rotate(tr[o].fa);
}
void access(int o) {
  for (int lst = 0; o; o = tr[lst = o].fa) {
    splay(o);
    tr[o].siz += tr[tr[o].child[1]].sum;
    tr[o].ad += tr[tr[o].child[1]].ans;
    tr[o].de += 1ll * tr[tr[o].child[1]].sum * tr[tr[o].child[1]].sum;
    tr[o].child[1] = lst;
    tr[o].siz -= tr[tr[o].child[1]].sum;
    tr[o].ad -= tr[tr[o].child[1]].ans;
    tr[o].de -= 1ll * tr[tr[o].child[1]].sum * tr[tr[o].child[1]].sum;
    pushup(o);
  }
}
void link(int x, int y) {
  access(y);
  splay(y);
  access(x);
  splay(x);
  tr[y].fa = x;
  tr[x].siz += tr[y].sum;
  tr[x].ad += tr[y].ans;
  tr[x].de += 1ll * tr[y].sum * tr[y].sum;
  return pushup(x);
}
void cut(int x, int y) {
  access(x), splay(x), splay(y);
  tr[x].siz -= tr[y].sum;
  tr[x].ad -= tr[y].ans;
  tr[x].de -= 1ll * tr[y].sum * tr[y].sum;
  tr[y].fa = 0;
  return pushup(x);
}
bool check(int x, int y) {
  access(y), splay(y), splay(x);
  return nr(y);
}
signed main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) scanf("%d", pre + i);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &tr[i].a);
    tr[i].ans = tr[i].all = tr[i].a;
    tr[i].siz = tr[i].sum = 1;
  }
  for (int i = 2; i <= n; ++i) link(pre[i], i);
  access(1), splay(1), Ans = tr[1].ans;
  printf("%.10lf\n", 1.0 * Ans / n / n);
  scanf("%d", &m);
  char op[10];
  for (int x, y; m--;) {
    scanf("%s%d%d", op, &x, &y);
    if (op[0] == 'P') {
      if (pre[y] == x || pre[x] == y) {
        printf("%.10lf\n", 1.0 * Ans / n / n);
        continue;
      }
      if (check(x, y)) swap(x, y);
      cut(pre[x], x), pre[x] = y, link(pre[x], x);
      access(1), splay(1), Ans = tr[1].ans;
    } else {
      access(x), splay(x), tr[x].a = y, pushup(x);
      access(1), splay(1), Ans = tr[1].ans;
    }
    printf("%.10lf\n", 1.0 * Ans / n / n);
  }
  return 0;
}