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 MAXN = 100005;
int n, m, far[MAXN], a[MAXN];
struct LCT {
  int fa[MAXN], ch[MAXN][2];
  int s[MAXN], xs[MAXN];
  long long xs2[MAXN], sum[MAXN], ans[MAXN], xans[MAXN];
  bool isroot(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
  int get(int x) { return ch[fa[x]][1] == x; }
  void up(int x) {
    s[x] = s[ch[x][0]] + s[ch[x][1]] + xs[x] + 1;
    sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + 1ll * a[x] * (xs[x] + 1);
    ans[x] = ans[ch[x][0]] + ans[ch[x][1]] + xans[x] +
             a[x] * (1ll * (xs[x] + 1) * (xs[x] + 1) - xs2[x]) +
             2ll * a[x] * (xs[x] + 1) * s[ch[x][1]] +
             2ll * sum[ch[x][0]] * (s[x] - s[ch[x][0]]);
  }
  void rotate(int x) {
    int f1 = fa[x], f2 = fa[f1], id = get(x);
    ch[f1][id] = ch[x][id ^ 1];
    fa[ch[f1][id]] = f1;
    if (!isroot(f1)) ch[f2][get(f1)] = x;
    fa[x] = f2;
    ch[x][id ^ 1] = f1;
    fa[f1] = x;
    up(f1);
  }
  void splay(int x) {
    while (!isroot(x)) {
      int f1 = fa[x];
      if (!isroot(f1)) {
        if (get(x) == get(f1))
          rotate(f1);
        else
          rotate(x);
      }
      rotate(x);
    }
    up(x);
  }
  void access(int x) {
    for (int y = 0; x; y = x, x = fa[x]) {
      splay(x);
      xs[x] += s[ch[x][1]] - s[y];
      xs2[x] += 1ll * s[ch[x][1]] * s[ch[x][1]] - 1ll * s[y] * s[y];
      xans[x] += ans[ch[x][1]] - ans[y];
      ch[x][1] = y;
      up(x);
    }
  }
  void link(int x, int y) {
    access(x), splay(x);
    access(y), splay(y);
    fa[x] = y;
    xs[y] += s[x];
    xans[y] += ans[x];
    xs2[y] += 1ll * s[x] * s[x];
    up(y);
  }
  void cut(int x, int y) {
    access(y), splay(y);
    splay(x);
    fa[x] = 0;
    xs[y] -= s[x];
    xans[y] -= ans[x];
    xs2[y] -= 1ll * s[x] * s[x];
    up(y);
  }
  bool is_anc(int x, int y) {
    access(y);
    splay(y), splay(x);
    return !isroot(y);
  }
  long long ask() {
    splay(1);
    return ans[1];
  }
  void modify(int x, int v) {
    access(x), splay(x);
    a[x] = v, up(x);
  }
} tr;
char op[2];
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) scanf("%d", &far[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) tr.s[i] = 1;
  for (int i = 2; i <= n; i++) tr.link(i, far[i]);
  printf("%.10lf\n", (double)tr.ask() / n / n);
  scanf("%d", &m);
  while (m--) {
    int x, y;
    scanf("%s%d%d", op, &x, &y);
    if (op[0] == 'P') {
      if (tr.is_anc(x, y)) swap(x, y);
      tr.cut(x, far[x]);
      far[x] = y;
      tr.link(x, far[x]);
    } else
      tr.modify(x, y);
    printf("%.10lf\n", (double)tr.ask() / n / n);
  }
  return 0;
}
