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 = 50010;
int n, m;
int a[maxn], p[maxn];
long long size[maxn], sum[maxn], ans[maxn], val[maxn], s1[maxn], s2[maxn];
int ch[maxn][2], fa[maxn];
template <typename Tp>
void read(Tp &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
bool isroot(int x) { return x != ch[fa[x]][0] && x != ch[fa[x]][1]; }
void pushup(int x) {
  sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + size[x];
  val[x] = val[ch[x][0]] + val[ch[x][1]] + a[x] * size[x];
  ans[x] = ans[ch[x][0]] + ans[ch[x][1]] + s1[x] +
           a[x] * (size[x] * size[x] - s2[x]) +
           2 * a[x] * size[x] * sum[ch[x][1]] +
           2 * val[ch[x][0]] * (sum[x] - sum[ch[x][0]]);
}
void update(int x, int y, int f) {
  size[x] += f * sum[y];
  s1[x] += f * ans[y];
  s2[x] += f * sum[y] * sum[y];
}
bool get(int x) { return x == ch[fa[x]][1]; }
void connect(int x, int f, int y) {
  ch[x][f] = y;
  fa[y] = x;
}
void rotate(int x) {
  int y = fa[x], z = fa[y], f = get(x);
  if (!isroot(y)) connect(z, y == ch[z][1], x);
  connect(y, f, ch[x][f ^ 1]);
  connect(x, f ^ 1, y);
  fa[x] = z;
  pushup(y), pushup(x);
}
void splay(int x) {
  for (int y = fa[x]; !isroot(x); rotate(x), y = fa[x])
    if (!isroot(y)) rotate(get(x) == get(y) ? y : x);
}
void access(int x) {
  for (int y = 0; x; y = x, x = fa[x]) {
    splay(x);
    update(x, ch[x][1], 1);
    ch[x][1] = y;
    update(x, y, -1);
    pushup(x);
  }
}
void link(int x, int y) {
  access(x);
  splay(x);
  splay(y);
  fa[y] = x;
  update(x, y, 1);
  pushup(x);
}
void cut(int x, int y) {
  access(x);
  splay(x);
  splay(y);
  update(x, y, -1);
  fa[y] = ch[x][1] = 0;
  pushup(x);
}
bool check(int x, int y) {
  access(y);
  splay(y);
  splay(x);
  return isroot(y);
}
void query() {
  access(1);
  splay(1);
  printf("%.10lf\n", (double)ans[1] / n / n);
}
int main() {
  read(n);
  for (int i = 2; i <= n; i++) read(p[i]);
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    size[i] = sum[i] = 1;
    ans[i] = val[i] = a[i];
  }
  for (int i = 2; i <= n; i++) link(p[i], i);
  query();
  read(m);
  for (int i = 1; i <= m; i++) {
    char op;
    int x, y;
    scanf("%c", &op);
    read(x), read(y);
    if (op == 'P') {
      if (p[x] == y || p[y] == x) {
        query();
        continue;
      }
      if (check(x, y)) {
        cut(p[x], x);
        link(y, x);
        p[x] = y;
      } else {
        cut(p[y], y);
        link(x, y);
        p[y] = x;
      }
    } else {
      access(x);
      splay(x);
      a[x] = y;
      pushup(x);
    }
    query();
  }
  return 0;
}