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 = 1e5 + 7;
struct data {
  int to, next;
} e[N];
int head[N], cnt;
void ins(int u, int v) {
  e[++cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
int c[N][2], fa[N], q[N], top, n, m;
double U;
long long notR[N], sumR[N], way[N], sumw[N], w[N], tag[N], ans;
bool isroot(int x) { return c[fa[x]][0] != x && c[fa[x]][1] != x; }
void add(int x, int val) {
  if (!x) return;
  tag[x] += val;
  way[x] += notR[x] * val;
}
void update(int x) {
  int l = c[x][0], r = c[x][1];
  sumR[x] = sumR[l] + sumR[r] + notR[x];
  sumw[x] = sumw[l] + sumw[r] + w[x] * notR[x];
}
void pushdown(int x) {
  int l = c[x][0], r = c[x][1];
  if (tag[x]) add(l, tag[x]), add(r, tag[x]), tag[x] = 0;
}
void rotate(int x) {
  int y = fa[x], z = fa[y], l, r;
  if (c[y][0] == x)
    l = 0;
  else
    l = 1;
  r = l ^ 1;
  if (!isroot(y)) c[z][c[z][1] == y] = x;
  fa[x] = z;
  fa[y] = x;
  fa[c[x][r]] = y;
  c[y][l] = c[x][r];
  c[x][r] = y;
  update(y);
  update(x);
}
void splay(int x) {
  q[++top] = x;
  for (int i = x; !isroot(i); i = fa[i]) q[++top] = fa[i];
  while (top) pushdown(q[top--]);
  while (!isroot(x)) {
    int y = fa[x], z = fa[y];
    if (!isroot(y)) {
      if (c[y][0] == x ^ c[z][0] == y)
        rotate(x);
      else
        rotate(y);
    }
    rotate(x);
  }
}
void access(int x) {
  for (int t = 0; x; t = x, x = fa[x])
    splay(x), notR[x] += sumR[c[x][1]], notR[x] -= sumR[t], c[x][1] = t,
                                                            update(x);
}
int pre(int x) {
  if (!c[x][0]) return fa[x];
  x = c[x][0];
  while (c[x][1]) x = c[x][1];
  return x;
}
void dfs(int x) {
  long long sub = 1, tot = 1;
  for (int i = head[x]; i; i = e[i].next) {
    dfs(e[i].to);
    tot += sub * notR[e[i].to] << 1;
    sub += notR[e[i].to];
  }
  ans += w[x] * tot;
  way[x] = tot;
  notR[x] = sub;
  update(x);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) scanf("%d", &fa[i]), ins(fa[i], i);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &w[i]);
  dfs(1);
  U = (long long)n * n;
  printf("%.12lf\n", ans / U);
  scanf("%d", &m);
  while (m--) {
    char op;
    int x, y;
    scanf(" %c%d%d", &op, &x, &y);
    if (op == 'P') {
      access(y);
      splay(x);
      if (!fa[x])
        swap(x, y);
      else
        access(x);
      int k = pre(x);
      int T = notR[x];
      if (k == y) goto over;
      access(k);
      splay(k);
      splay(x);
      notR[k] -= T;
      update(k);
      add(k, -(T << 1));
      ans -= sumw[k] * T << 1;
      access(y);
      splay(y);
      ans += sumw[y] * T << 1;
      add(y, T << 1);
      notR[y] += T;
      update(y);
      fa[x] = y;
    } else
      splay(x), ans += way[x] * (y - w[x]), w[x] = y, update(x);
  over:;
    printf("%.12lf\n", ans / U);
  }
  return 0;
}
