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 maxn = 50005;
int n, q;
char c[2];
long long ans;
namespace Lct {
int ch[maxn][2], f[maxn], tag[maxn], val[maxn], sz[maxn], vsz[maxn];
long long cnt[maxn], sum[maxn];
void init() {
  for (int i = 1; i <= n; i++) cnt[i] = sz[i] = vsz[i] = 1;
}
bool isroot(int x) { return ch[f[x]][0] != x && ch[f[x]][1] != x; }
void apply(int x, int _tag) {
  cnt[x] += 2LL * vsz[x] * _tag;
  sz[x] += _tag;
  tag[x] += _tag;
}
void pushdown(int x) {
  if (tag[x]) {
    if (ch[x][0]) apply(ch[x][0], tag[x]);
    if (ch[x][1]) apply(ch[x][1], tag[x]);
    tag[x] = 0;
  }
}
void pushup(int x) {
  sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + 1LL * val[x] * vsz[x];
}
void rotate(int x) {
  int y = f[x], z = f[y], d = ch[y][1] == x;
  if (!isroot(y)) ch[z][ch[z][1] == y] = x;
  f[x] = z;
  f[y] = x;
  f[ch[x][!d]] = y;
  ch[y][d] = ch[x][!d];
  ch[x][!d] = y;
  pushup(y);
  pushup(x);
}
void push(int x) {
  if (!isroot(x)) push(f[x]);
  pushdown(x);
}
void splay(int x) {
  push(x);
  for (int y = f[x], z = f[y]; !isroot(x); y = f[x], z = f[y]) {
    if (!isroot(y)) ((ch[z][1] == y) ^ (ch[y][1] == x)) ? rotate(x) : rotate(y);
    rotate(x);
  }
}
void access(int x) {
  int y = 0, tmp = x;
  while (x) {
    splay(x);
    if (y) {
      while (ch[y][0]) y = ch[y][0];
      splay(y);
    }
    vsz[x] = sz[x] - sz[y];
    ch[x][1] = y;
    pushup(x);
    y = x;
    x = f[x];
  }
  splay(tmp);
}
bool anc(int x, int y) {
  access(y);
  splay(x);
  return f[y] > 0;
}
void link(int x, int y) {
  access(x);
  access(y);
  ans += 2LL * sz[x] * sum[y];
  apply(y, sz[x]);
  f[x] = y;
}
void cut(int x) {
  access(x);
  ans -= 2LL * sz[x] * sum[ch[x][0]];
  if (ch[x][0]) apply(ch[x][0], -sz[x]);
  f[ch[x][0]] = 0;
  ch[x][0] = 0;
  pushup(x);
}
void modify(int x, int v) {
  splay(x);
  ans += 1LL * (v - val[x]) * cnt[x];
  val[x] = v;
  pushup(x);
}
}  // namespace Lct
int main() {
  scanf("%d", &n);
  Lct::init();
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    Lct::link(i, p);
  }
  for (int i = 1; i <= n; i++) {
    int v;
    scanf("%d", &v);
    Lct::modify(i, v);
  }
  printf("%.10lf\n", double(ans) / double(n) / double(n));
  scanf("%d", &q);
  while (q--) {
    scanf("%s", c);
    if (c[0] == 'P') {
      int x, y;
      scanf("%d%d", &x, &y);
      if (Lct::anc(x, y)) swap(x, y);
      Lct::cut(x);
      Lct::link(x, y);
    } else {
      int x, v;
      scanf("%d%d", &x, &v);
      Lct::modify(x, v);
    }
    printf("%.10lf\n", double(ans) / double(n) / double(n));
  }
  return 0;
}