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 = 100005;
int n, m, x, y, f[N], fa[N], ch[N][2];
char op[5];
long long a[N], siz[N], ssiz[N], ans[N], sans[N], sqsiz[N], all[N];
bool isroot(int u) { return ch[fa[u]][0] != u && ch[fa[u]][1] != u; }
int which(int u) { return u == ch[fa[u]][1]; }
void pushup(int o) {
  int lc = ch[o][0], rc = ch[o][1];
  siz[o] = siz[lc] + siz[rc] + ssiz[o];
  all[o] = all[lc] + all[rc] + a[o] * ssiz[o];
  ans[o] = ans[lc] + ans[rc] + sans[o];
  ans[o] += (ssiz[o] * ssiz[o] - sqsiz[o]) * a[o];
  ans[o] += 2 * siz[rc] * ssiz[o] * a[o];
  ans[o] += 2 * all[lc] * (siz[o] - siz[lc]);
}
void rotate(int x) {
  int y = fa[x], z = fa[y], md = which(x);
  if (!isroot(y)) {
    ch[z][which(y)] = x;
  }
  fa[x] = z;
  ch[y][md] = ch[x][!md];
  fa[ch[y][md]] = y;
  ch[x][!md] = y;
  fa[y] = x;
  pushup(y);
  pushup(x);
}
void splay(int u) {
  while (!isroot(u)) {
    if (!isroot(fa[u])) {
      rotate(which(u) == which(fa[u]) ? fa[u] : u);
    }
    rotate(u);
  }
}
void access(int u) {
  for (int v = 0; u; v = u, u = fa[u]) {
    splay(u);
    int rc = ch[u][1];
    ssiz[u] += siz[rc];
    sans[u] += ans[rc];
    sqsiz[u] += siz[rc] * siz[rc];
    ch[u][1] = v;
    ssiz[u] -= siz[v];
    sans[u] -= ans[v];
    sqsiz[u] -= siz[v] * siz[v];
    pushup(u);
  }
}
void link(int u, int v) {
  access(v);
  splay(v);
  access(u);
  splay(u);
  fa[u] = v;
  ssiz[v] += siz[u];
  sans[v] += ans[u];
  sqsiz[v] += siz[u] * siz[u];
  pushup(v);
}
void cut(int u, int v) {
  access(u);
  access(v);
  splay(v);
  splay(u);
  fa[u] = 0;
  ssiz[v] -= siz[u];
  sans[v] -= ans[u];
  sqsiz[v] -= siz[u] * siz[u];
  pushup(v);
}
bool check(int u, int v) {
  access(v);
  splay(v);
  splay(u);
  if (isroot(v)) {
    return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &f[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    ans[i] = all[i] = a[i];
  }
  for (int i = 1; i <= n; i++) {
    siz[i] = 1;
    ssiz[i] = 1;
  }
  for (int i = 2; i <= n; i++) {
    link(i, f[i]);
  }
  access(1);
  splay(1);
  printf("%.10lf\n", 1.0 * ans[1] / n / n);
  scanf("%d", &m);
  while (m--) {
    scanf("%s%d%d", op, &x, &y);
    if (op[0] == 'P') {
      if (check(x, y)) {
        swap(x, y);
      }
      cut(x, f[x]);
      f[x] = y;
      link(x, f[x]);
      access(1);
      splay(1);
      printf("%.10lf\n", 1.0 * ans[1] / n / n);
    } else {
      access(x);
      splay(x);
      a[x] = y;
      pushup(x);
      printf("%.10lf\n", 1.0 * ans[x] / n / n);
    }
  }
  return 0;
}
