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 std::cerr;
using std::endl;
const int N = 1e5 + 233;
int n, q, val[N], mem[N];
namespace LCT {
int fa[N], ch[N][2];
long long A[N], B[N], C[N], D[N], E[N], F[N];
inline bool isroot(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
inline void pushup(int x) {
  E[x] = E[ch[x][0]] + E[ch[x][1]] + B[x];
  F[x] = F[ch[x][0]] + F[ch[x][1]] + B[x] * val[x];
  A[x] = A[ch[x][0]] + A[ch[x][1]] + D[x] + (B[x] * B[x] - C[x]) * val[x] +
         E[ch[x][1]] * B[x] * val[x] * 2 +
         (E[x] - E[ch[x][0]]) * F[ch[x][0]] * 2;
}
inline void rotate(int x) {
  int y = fa[x], z = fa[y], k = ch[y][1] == x;
  if (!isroot(y)) ch[z][ch[z][1] == y] = x;
  fa[x] = z, ch[y][k] = ch[x][k ^ 1];
  fa[ch[y][k]] = y;
  ch[x][k ^ 1] = y;
  fa[y] = x;
  pushup(y), pushup(x);
}
inline void splay(int x) {
  while (!isroot(x)) {
    int y = fa[x], z = fa[y];
    if (!isroot(y)) rotate((ch[z][1] == y) != (ch[y][1] == x) ? x : y);
    rotate(x);
  }
}
inline void access(int x) {
  for (int y = 0; x; x = fa[y = x]) {
    splay(x);
    B[x] += E[ch[x][1]];
    C[x] += E[ch[x][1]] * E[ch[x][1]];
    D[x] += A[ch[x][1]];
    ch[x][1] = y;
    B[x] -= E[ch[x][1]];
    C[x] -= E[ch[x][1]] * E[ch[x][1]];
    D[x] -= A[ch[x][1]];
    pushup(x);
  }
}
inline void link(int x, int y) {
  access(x), splay(x);
  access(y), splay(y);
  fa[x] = y;
  B[y] += E[x];
  C[y] += E[x] * E[x];
  D[y] += A[x];
  pushup(y);
}
inline void cut(int x, int y) {
  access(y), splay(y), splay(x);
  B[y] -= E[x];
  C[y] -= E[x] * E[x];
  D[y] -= A[x];
  fa[x] = 0;
  pushup(y);
}
inline int check(int x, int y) {
  access(x);
  splay(x);
  splay(y);
  return !isroot(x);
}
inline long double calc() {
  access(1), splay(1);
  long double ret = A[1];
  ret = ret / n / n;
  return ret;
}
}  // namespace LCT
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  std::cin >> n;
  for (int i = 2; i <= n; ++i) std::cin >> mem[i];
  for (int i = 1; i <= n; ++i) {
    std::cin >> val[i];
    LCT::A[i] = LCT::F[i] = val[i];
    LCT::B[i] = LCT::E[i] = 1;
  }
  for (int i = 2; i <= n; ++i) {
    LCT::link(i, mem[i]);
  }
  std::cout << std::fixed << std::setprecision(12) << LCT::calc() << '\n';
  std::cin >> q;
  while (q--) {
    char op;
    int x, y;
    std::cin >> op >> x >> y;
    if (op == 'P') {
      if (LCT::check(y, x)) std::swap(x, y);
      LCT::cut(x, mem[x]);
      mem[x] = y;
      LCT::link(x, y);
    } else {
      LCT::access(x);
      LCT::splay(x);
      val[x] = y;
      LCT::pushup(x);
    }
    std::cout << std::fixed << std::setprecision(12) << LCT::calc() << '\n';
  }
  return 0;
}