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;
int n, fa[100000 + 5], root, ch[100000 + 5][2], rev[100000 + 5];
int siz[100000 + 5], sizz[100000 + 5];
long long sum[100000 + 5], ans[100000 + 5];
long long add[100000 + 5], del[100000 + 5];
int a[100000 + 5], pre[100000 + 5];
int son(int x) { return (ch[fa[x]][1] == x); }
bool isroot(int x) { return (ch[fa[x]][0] != x) && (ch[fa[x]][1] != x); }
void pushdown(int pos) {
  if (rev[pos]) {
    rev[pos] = 0;
    rev[ch[pos][0]] ^= 1;
    swap(ch[ch[pos][0]][0], ch[ch[pos][0]][1]);
    rev[ch[pos][1]] ^= 1;
    swap(ch[ch[pos][1]][0], ch[ch[pos][1]][1]);
  }
}
void pushup(int pos) {
  if (!pos) return;
  siz[pos] = siz[ch[pos][0]] + siz[ch[pos][1]] + sizz[pos];
  sum[pos] = sum[ch[pos][0]] + sum[ch[pos][1]] + 1ll * a[pos] * sizz[pos];
  ans[pos] = ans[ch[pos][0]] + ans[ch[pos][1]] + add[pos] +
             1ll * a[pos] * (1ll * sizz[pos] * sizz[pos] - del[pos]) +
             2ll * a[pos] * sizz[pos] * siz[ch[pos][1]] +
             2ll * sum[ch[pos][0]] * (siz[pos] - siz[ch[pos][0]]);
}
void rotate(int x) {
  int y = fa[x], z = fa[y], b = son(x), c = son(y), a = ch[x][!b];
  if (!isroot(y)) ch[z][c] = x;
  fa[x] = z;
  if (a) fa[a] = y;
  ch[y][b] = a;
  ch[x][!b] = y;
  fa[y] = x;
  pushup(y);
  pushup(x);
}
void cleartag(int pos) {
  if (isroot(pos)) {
    pushdown(pos);
    return;
  }
  cleartag(fa[pos]);
  pushdown(pos);
}
void splay(int x) {
  cleartag(x);
  while (!isroot(x)) {
    int y = fa[x], z = fa[y];
    if (isroot(y))
      rotate(x);
    else {
      if (son(x) == son(y))
        rotate(y), rotate(x);
      else
        rotate(x), rotate(x);
    }
  }
  pushup(x);
}
void access(int x) {
  int y = 0, tmp = x;
  while (x) {
    splay(x);
    sizz[x] += siz[ch[x][1]];
    del[x] += 1ll * siz[ch[x][1]] * siz[ch[x][1]];
    add[x] += ans[ch[x][1]];
    ch[x][1] = y;
    sizz[x] -= siz[ch[x][1]];
    del[x] -= 1ll * siz[ch[x][1]] * siz[ch[x][1]];
    add[x] -= ans[ch[x][1]];
    y = x;
    pushup(x);
    x = fa[x];
  }
  splay(tmp);
}
void link(int x, int y) {
  access(y);
  access(x);
  fa[y] = x;
  sizz[x] += siz[y];
  add[x] += ans[y];
  del[x] += 1ll * siz[y] * siz[y];
  pushup(x);
}
void cut(int x, int y) {
  access(x);
  splay(y);
  sizz[x] -= siz[y];
  add[x] -= ans[y];
  del[x] -= 1ll * siz[y] * siz[y];
  fa[y] = 0;
  pushup(x);
}
char op[10];
int main() {
  int n, m, i, j;
  scanf("%d", &n);
  for (i = 2; i <= n; i++) scanf("%d", &pre[i]);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    siz[i] = sizz[i] = 1;
    ans[i] = sum[i] = a[i];
  }
  for (i = 2; i <= n; i++) link(pre[i], i);
  access(1);
  printf("%.10lf\n", double(ans[1]) / n / n);
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    int x, y;
    scanf("%s%d%d", op, &x, &y);
    if (op[0] == 'P') {
      if (pre[x] != y && pre[y] != x) {
        access(y);
        splay(x);
        if (!isroot(y)) swap(x, y);
        cut(pre[x], x);
        pre[x] = y;
        link(y, x);
      }
      access(1);
      printf("%.10lf\n", double(ans[1]) / n / n);
    } else {
      access(x);
      a[x] = y;
      pushup(x);
      printf("%.10lf\n", double(ans[x]) / n / n);
    }
  }
}
