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 = 1.5e5 + 7;
int n, m;
int p[N], f[N], ch[N][2], a[N], sz[N], sum[N];
long long add[N], sub[N], ans[N], val[N];
inline bool Son(int x) { return ch[f[x]][1] == x; }
inline bool isroot(int x) { return ch[f[x]][0] != x && ch[f[x]][1] != x; }
inline void update(int x) {
  sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + sz[x];
  val[x] = val[ch[x][0]] + val[ch[x][1]] + 1ll * a[x] * sz[x];
  ans[x] = ans[ch[x][0]] + ans[ch[x][1]] + add[x] +
           (1ll * sz[x] * sz[x] + 2ll * sz[x] * sum[ch[x][1]] - sub[x]) * a[x] +
           2ll * val[ch[x][0]] * (sz[x] + sum[ch[x][1]]);
}
inline void rotate(int x) {
  int fa = f[x], g = f[fa], k = Son(x);
  if (!isroot(fa)) ch[g][Son(fa)] = x;
  f[x] = g;
  ch[fa][k] = ch[x][k ^ 1];
  f[ch[fa][k]] = fa;
  ch[x][k ^ 1] = fa;
  f[fa] = x;
  update(fa);
  update(x);
}
inline void splay(int x) {
  for (; !isroot(x); rotate(x))
    if (!isroot(f[x])) rotate(Son(x) == Son(f[x]) ? f[x] : x);
}
inline void access(int x) {
  for (int y = 0; x; y = x, x = f[x]) {
    splay(x);
    if (ch[x][1]) {
      sub[x] += 1ll * sum[ch[x][1]] * sum[ch[x][1]];
      add[x] += ans[ch[x][1]];
      sz[x] += sum[ch[x][1]];
    }
    ch[x][1] = y;
    if (y) {
      sub[x] -= 1ll * sum[y] * sum[y];
      add[x] -= ans[y];
      sz[x] -= sum[y];
    }
    update(x);
  }
}
inline int findroot(int x) {
  splay(x);
  while (ch[x][0]) x = ch[x][0];
  return splay(x), x;
}
inline void link(int fa, int x) {
  access(fa);
  splay(fa);
  splay(x);
  f[x] = fa;
  sub[fa] += 1ll * sum[x] * sum[x];
  add[fa] += ans[x];
  sz[fa] += sum[x];
  update(fa);
}
inline void cut(int fa, int x) {
  access(fa);
  splay(fa);
  splay(x);
  f[x] = 0;
  sub[fa] -= 1ll * sum[x] * sum[x];
  add[fa] -= ans[x];
  sz[fa] -= sum[x];
  update(fa);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) scanf("%d", &p[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) sz[i] = sum[i] = 1, val[i] = ans[i] = a[i];
  for (int i = 2; i <= n; ++i) link(p[i], i);
  scanf("%d", &m);
  splay(1);
  printf("%.10lf\n", (double)(1.0l * ans[1] / n / n));
  for (int i = 1, x, y; i <= m; ++i) {
    char ch = getchar();
    while (!isalpha(ch)) ch = getchar();
    scanf("%d%d", &x, &y);
    if (ch == 'V')
      access(x), splay(x), a[x] = y, update(x);
    else {
      access(y);
      if (findroot(x) == 1) {
        access(p[y]);
        cut(p[y], y);
        p[y] = x;
        link(p[y], y);
      } else {
        access(p[x]);
        cut(p[x], x);
        p[x] = y;
        link(p[x], x);
      }
    }
    splay(1);
    printf("%.10lf\n", (double)(1.0l * ans[1] / n / n));
  }
  return 0;
}
