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;
struct edge {
  int v, nxt;
} e[500005];
int n, m, val[100005], sz[100005], szi[100005], f[100005], ch[100005][2];
int rev[100005], p[100005], lazy[100005], h[100005], t;
long long sum[100005], ans, c[100005];
void add(int u, int v) {
  e[++t].v = v;
  e[t].nxt = h[u];
  h[u] = t;
}
void pushup(int x) {
  sz[x] = 1 + sz[ch[x][0]] + sz[ch[x][1]] + szi[x];
  sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + 1ll * val[x] * (szi[x] + 1);
}
bool nroot(int x) { return ch[f[x]][0] == x || ch[f[x]][1] == x; }
void rotate(int x) {
  int y = f[x], z = f[y], k = ch[y][1] == x, r = ch[x][!k];
  if (nroot(y)) ch[z][ch[z][1] == y] = x;
  ch[x][!k] = y;
  ch[y][k] = r;
  f[r] = y;
  f[y] = x;
  f[x] = z;
  pushup(y);
  pushup(x);
}
void pushr(int x) {
  rev[x] ^= 1;
  swap(ch[x][0], ch[x][1]);
}
void col(int x, int v) {
  c[x] += 2ll * v * (szi[x] + 1);
  lazy[x] += v;
}
void pushdown(int x) {
  if (rev[x]) {
    if (ch[x][0]) pushr(ch[x][0]);
    if (ch[x][1]) pushr(ch[x][1]);
    rev[x] = 0;
  }
  if (lazy[x]) {
    if (ch[x][0]) col(ch[x][0], lazy[x]);
    if (ch[x][1]) col(ch[x][1], lazy[x]);
    lazy[x] = 0;
  }
}
void pushall(int x) {
  if (nroot(x)) pushall(f[x]);
  pushdown(x);
}
void splay(int x) {
  pushall(x);
  while (nroot(x)) {
    int y = f[x], z = f[y];
    if (nroot(y)) rotate((ch[z][1] == y) ^ (ch[y][1] == x) ? x : y);
    rotate(x);
  }
}
void access(int x) {
  for (int y = 0; x; y = x, x = f[x]) {
    splay(x);
    szi[x] = szi[x] + sz[ch[x][1]] - sz[y];
    ch[x][1] = y;
    pushup(x);
  }
}
void makeroot(int x) {
  access(x);
  splay(x);
  pushr(x);
}
int findroot(int x) {
  access(x);
  splay(x);
  while (ch[x][0]) pushdown(x), x = ch[x][0];
  splay(x);
  return x;
}
void split(int x, int y) {
  makeroot(x);
  access(y);
  splay(y);
}
void link(int x, int y) {
  split(x, y);
  if (!f[x] && x != y) {
    f[x] = y;
    szi[y] += sz[x];
    pushup(y);
  }
}
void cut(int x, int y) {
  split(x, y);
  if (f[x] == y && ch[y][0] == x && !ch[x][1]) {
    ch[y][0] = f[x] = 0;
    pushup(y);
  }
}
void tlink(int x) {
  int y = p[x];
  split(1, y);
  col(y, sz[x]);
  makeroot(x);
  ans += 2ll * sz[x] * sum[y];
  link(x, y);
}
void tcut(int x) {
  int y = p[x];
  cut(x, y);
  split(1, y);
  col(y, -sz[x]);
  makeroot(x);
  ans -= 2ll * sz[x] * sum[y];
}
void modify(int x, int v) {
  split(1, x);
  ans += (v - val[x]) * c[x];
  val[x] = v;
  pushup(x);
}
void dfs(int u, int fa) {
  if (fa) tlink(u);
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) sz[i] = 1, c[i] = 1;
  for (int i = 2; i <= n; i++) scanf("%d", &p[i]), add(i, p[i]), add(p[i], i);
  for (int i = 1; i <= n; i++) scanf("%d", &val[i]), ans += val[i];
  dfs(1, 0);
  scanf("%d", &m);
  printf("%.10f\n", 1.0 * ans / n / n);
  for (int i = 1; i <= m; i++) {
    int x, y;
    char op;
    scanf(" %c%d%d", &op, &x, &y);
    if (op == 'P') {
      split(1, y);
      splay(x);
      int nw = x;
      while (ch[nw][0]) pushdown(nw), nw = ch[nw][0];
      if (nw == 1) swap(x, y);
      tcut(x);
      p[x] = y;
      tlink(x);
    } else {
      modify(x, y);
    }
    printf("%.10f\n", 1.0 * ans / n / n);
  }
  return 0;
}
