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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using std::cin;
using std::cout;
const int maxn = 100100;
int n, m;
unsigned w[maxn];
int a[maxn], aw[maxn];
int size[maxn], son[maxn], dfn[maxn];
int fa[maxn];
int head[maxn], next[maxn];
template <class ptr>
inline void tr(ptr a) {
  static int b[maxn];
  memcpy(b + 1, a + 1, n << 2);
  for (int i = 1; i <= n; ++i) a[dfn[i]] = b[i];
}
inline void link(int fa, int x) { next[x] = head[fa], head[fa] = x; }
inline void dfs0(int x) {
  size[x] = 1;
  for (int i = head[x]; i; i = next[i]) {
    dfs0(i);
    size[x] += size[i];
    if (size[i] > size[son[x]]) son[x] = i;
  }
  w[x] = (unsigned)size[x] * size[x];
}
inline void dfs1(int x) {
  static int tot;
  dfn[x] = ++tot;
  if (son[x]) dfs1(son[x]);
  for (int i = head[x]; i; i = next[i])
    if (i != son[x]) dfs1(i);
}
long long su = 0;
inline double calc() { return (double)su / n / n; }
inline void add(int x, int v) {
  while (x) {
    su += (2ll * size[x] * v + (long long)v * v) * aw[x];
    size[x] += v;
    w[x] = size[x] * size[x];
    x = fa[x];
  }
}
inline void change(int x, int y) {
  add(fa[x], -size[x]);
  add(y, size[x]);
  fa[x] = y;
  su -= (long long)w[x] * aw[x];
  aw[x] = a[x] - a[fa[x]];
  su += (long long)w[x] * aw[x];
}
inline bool is_anc(int x, int y) {
  while (y) {
    if (x == y) return 1;
    y = fa[y];
  }
  return 0;
}
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    cin >> fa[i];
    link(fa[i], i);
  }
  dfs0(1);
  dfs1(1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    aw[i] = a[i] - a[fa[i]];
  }
  for (int i = 1; i <= n; ++i) su += (long long)w[i] * aw[i];
  for (int i = 1; i <= n; ++i) {
    fa[i] = dfn[fa[i]];
  }
  tr(fa);
  tr(a);
  tr(w);
  tr(aw);
  tr(size);
  printf("%.10lf\n", calc());
  cin >> m;
  for (int i = 1, x, y; i <= m; ++i) {
    char ch;
    cin >> ch >> x >> y;
    x = dfn[x];
    if (ch == 'P') {
      y = dfn[y];
      if (is_anc(x, y)) {
        change(y, x);
      } else {
        change(x, y);
      }
    } else {
      const int delt = y - a[x];
      a[x] += delt, aw[x] += delt, su += (long long)w[x] * delt;
      long long sm = 0;
      for (int i = 1; i <= n; ++i)
        if (fa[i] == x) sm += w[i], aw[i] -= delt;
      su -= sm * delt;
    }
    printf("%.10lf\n", calc());
  }
}