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 N = 1e5 + 10;
int n, m, a[N], fa[N];
char t[11];
struct LCT {
  int f[N], ch[N][2];
  long long sum[N], size[N], all[N], ad[N], de[N], ans[N];
  bool isroot(int x) { return ch[f[x]][0] != x && ch[f[x]][1] != x; }
  bool get(int x) { return ch[f[x]][1] == x; }
  void pushup(int x) {
    sum[x] = size[x] + sum[ch[x][0]] + sum[ch[x][1]];
    all[x] = all[ch[x][0]] + all[ch[x][1]] + 1ll * a[x] * size[x];
    ans[x] = ans[ch[x][0]] + ans[ch[x][1]] + ad[x] +
             a[x] * (size[x] * size[x] - de[x]) +
             2ll * a[x] * size[x] * sum[ch[x][1]] +
             2ll * all[ch[x][0]] * (sum[x] - sum[ch[x][0]]);
  }
  void rotate(int x) {
    int y = f[x], z = f[y], k = get(x), l = ch[x][k ^ 1];
    if (!isroot(y)) ch[z][ch[z][1] == y] = x;
    f[x] = z;
    ch[x][k ^ 1] = y;
    f[y] = x;
    ch[y][k] = l;
    if (l) f[l] = y;
    pushup(y), pushup(x);
  }
  void splay(int x) {
    for (int fa; !isroot(x); rotate(x)) {
      if (!isroot(fa = f[x])) {
        rotate(get(x) == get(fa) ? fa : x);
      }
    }
  }
  void access(int x) {
    for (int y = 0; x; y = x, x = f[x]) {
      splay(x);
      size[x] += sum[ch[x][1]];
      ad[x] += ans[ch[x][1]];
      de[x] += sum[ch[x][1]] * sum[ch[x][1]];
      ch[x][1] = y;
      size[x] -= sum[ch[x][1]];
      ad[x] -= ans[ch[x][1]];
      de[x] -= sum[ch[x][1]] * sum[ch[x][1]];
      pushup(x);
    }
  }
  void link(int x, int y) {
    access(y), splay(y);
    access(x), splay(x);
    f[y] = x;
    size[x] += sum[y];
    ad[x] += ans[y];
    de[x] += sum[y] * sum[y];
    pushup(x);
  }
  void cut(int x, int y) {
    access(x), splay(x);
    splay(y);
    size[x] -= sum[y];
    ad[x] -= ans[y];
    de[x] -= sum[y] * sum[y];
    f[y] = 0;
    pushup(x);
  }
  bool check(int x, int y) {
    access(y);
    splay(x);
    return !f[x];
  }
} L;
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) scanf("%d", &fa[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    L.sum[i] = L.size[i] = 1;
    L.ans[i] = L.all[i] = a[i];
  }
  for (int i = 2; i <= n; i++) {
    L.link(fa[i], i);
  }
  L.access(1);
  L.splay(1);
  printf("%0.10lf\n", (double)L.ans[1] / n / n);
  scanf("%d", &m);
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%s%d%d", t + 1, &x, &y);
    if (t[1] == 'P') {
      if (L.check(x, y)) {
        L.cut(fa[y], y);
        L.link(x, y);
        fa[y] = x;
      } else {
        L.cut(fa[x], x);
        L.link(y, x);
        fa[x] = y;
      }
    } else {
      L.access(x);
      L.splay(x);
      a[x] = y;
      L.pushup(x);
    }
    L.access(1);
    L.splay(1);
    printf("%0.10lf\n", (double)L.ans[1] / n / n);
  }
  return 0;
}