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 = 5e4 + 9;
int f[N], son[N][2];
long long val[N], sum[N], a[N], sum2[N];
int vir_sz[N], sz[N];
long long vir_sum[N], sz2[N];
bool isroot(int x) { return !f[x] || son[f[x]][0] != x && son[f[x]][1] != x; }
void up(int x) {
  sz[x] = vir_sz[x] + sz[son[x][0]] + sz[son[x][1]];
  sum[x] = sum[son[x][0]] + sum[son[x][1]] + vir_sum[x] +
           2 * val[son[x][0]] * (sz[x] - sz[son[x][0]]) +
           a[x] * (2ll * sz[son[x][1]] * vir_sz[x] +
                   1ll * vir_sz[x] * vir_sz[x] - sz2[x]);
  val[x] = val[son[x][0]] + val[son[x][1]] + 1ll * a[x] * vir_sz[x];
}
void vir_up(int x, int y, int v) {
  vir_sz[x] += v * sz[y];
  sz2[x] += 1ll * v * sz[y] * sz[y];
  vir_sum[x] += v * sum[y];
}
void rotate(int x) {
  int y = f[x], w = son[y][1] == x;
  son[y][w] = son[x][w ^ 1];
  if (son[x][w ^ 1]) f[son[x][w ^ 1]] = y;
  if (f[y]) {
    int z = f[y];
    if (son[z][0] == y)
      son[z][0] = x;
    else if (son[z][1] == y)
      son[z][1] = x;
  }
  f[x] = f[y];
  f[y] = x, son[x][w ^ 1] = y;
  up(y), up(x);
}
void splay(int x) {
  int s = 1, y;
  while (!isroot(x)) {
    y = f[x];
    if (!isroot(y)) {
      if ((son[f[y]][0] == y) ^ (son[y][0] == x))
        rotate(x);
      else
        rotate(y);
    }
    rotate(x);
  }
  up(x);
}
void access(int x) {
  for (int y = 0; x; y = x, x = f[x])
    splay(x), vir_up(x, son[x][1], 1), son[x][1] = y, vir_up(x, y, -1), up(x);
}
void makeroot(int x) {
  access(x);
  splay(x);
}
void link(int x, int y) {
  makeroot(x);
  makeroot(y);
  f[x] = y;
  vir_up(y, x, 1);
  access(x);
}
void cut(int x, int y) {
  makeroot(x);
  splay(y);
  f[x] = 0;
  son[y][1] = 0;
  up(y);
}
int fa[N];
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = (2); i < (n + 1); ++i) scanf("%d", &fa[i]);
  for (int i = (1); i < (n + 1); ++i)
    scanf("%lld", &a[i]), sz[i] = vir_sz[i] = 1, val[i] = sum[i] = a[i];
  for (int i = (2); i < (n + 1); ++i) link(i, fa[i]);
  access(1);
  splay(1);
  printf("%.9lf\n", 1.0 * sum[1] / n / n);
  scanf("%d", &m);
  while (m--) {
    char str[3];
    scanf("%s", str);
    int u, v;
    scanf("%d%d", &u, &v);
    if (str[0] == 'P') {
      swap(u, v);
      access(u);
      splay(u);
      splay(v);
      if (!isroot(u)) {
        cut(u, fa[u]);
        link(u, v);
        fa[u] = v;
      } else {
        cut(v, fa[v]);
        link(v, u);
        fa[v] = u;
      }
    } else {
      access(u);
      splay(u);
      a[u] = v;
      up(u);
    }
    access(1);
    splay(1);
    printf("%.9lf\n", 1.0 * sum[1] / n / n);
  }
}
