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 = 5e4 + 10;
int n, m, a[N], f[N];
double Ans;
struct node {
  int ch[2], fa, si, sz;
  long long ans, all, sim, sum;
} t[N];
char c[5];
bool isroot(int x) { return t[t[x].fa].ch[0] != x && t[t[x].fa].ch[1] != x; }
void pushup(int x) {
  int l = t[x].ch[0], r = t[x].ch[1];
  t[x].sz = t[l].sz + t[r].sz + t[x].si;
  t[x].all = 1ll * a[x] * t[x].si + t[l].all + t[r].all;
  t[x].ans = t[l].ans + t[r].ans + t[x].sum +
             1ll * a[x] * (1ll * t[x].si * t[x].si - t[x].sim) +
             2ll * a[x] * t[x].si * t[r].sz +
             2ll * t[l].all * (t[x].sz - t[l].sz);
}
void rotate(int x) {
  int y = t[x].fa, z = t[y].fa;
  int k = (t[y].ch[1] == x);
  t[x].fa = z;
  if (!isroot(y)) t[z].ch[t[z].ch[1] == y] = x;
  t[y].ch[k] = t[x].ch[k ^ 1];
  t[t[x].ch[k ^ 1]].fa = y;
  t[x].ch[k ^ 1] = y;
  t[y].fa = x;
  pushup(y);
}
void splay(int x) {
  while (!isroot(x)) {
    int y = t[x].fa, z = t[y].fa;
    if (!isroot(y)) {
      (t[y].ch[0] == x) ^ (t[z].ch[0] == y) ? rotate(x) : rotate(y);
    }
    rotate(x);
  }
  pushup(x);
}
void access(int x) {
  for (int y = 0; x; y = x, x = t[x].fa) {
    splay(x);
    int z = t[x].ch[1];
    t[x].si += t[z].sz;
    t[x].sim += 1ll * t[z].sz * t[z].sz;
    t[x].sum += t[z].ans;
    t[x].ch[1] = y;
    t[x].si -= t[y].sz;
    t[x].sim -= 1ll * t[y].sz * t[y].sz;
    t[x].sum -= t[y].ans;
    pushup(x);
  }
}
void Link(int x, int y) {
  access(x), splay(x);
  access(y), splay(y);
  t[y].fa = x;
  t[x].si += t[y].sz;
  t[x].sim += 1ll * t[y].sz * t[y].sz;
  t[x].sum += t[y].ans;
  pushup(x);
}
void Cut(int x, int y) {
  access(x), splay(x);
  splay(y);
  t[y].fa = 0;
  t[x].si -= t[y].sz;
  t[x].sim -= 1ll * t[y].sz * t[y].sz;
  t[x].sum -= t[y].ans;
  pushup(x);
}
bool check(int x, int y) {
  access(y);
  splay(y);
  splay(x);
  return !isroot(y);
}
int main() {
  scanf("%d", &n);
  for (int i = 2, x; i <= n; i++) {
    scanf("%d", &f[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    t[i].ans = t[i].all = a[i];
    t[i].si = t[i].sz = 1;
  }
  for (int i = 2; i <= n; i++) Link(f[i], i);
  access(1), splay(1);
  Ans = t[1].ans;
  printf("%.10lf\n", Ans / n / n);
  scanf("%d", &m);
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%s", c);
    scanf("%d%d", &x, &y);
    if (c[0] == 'P') {
      if (f[x] == y || f[y] == x) {
        printf("%.10lf\n", Ans / n / n);
        continue;
      }
      if (check(x, y)) swap(x, y);
      Cut(f[x], x);
      f[x] = y;
      Link(y, x);
      access(1), splay(1);
      Ans = t[1].ans;
    } else {
      access(x), splay(x);
      a[x] = y;
      pushup(x);
      Ans = t[x].ans;
    }
    printf("%.10lf\n", Ans / n / n);
  }
  return 0;
}
