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, m, a[100005], F[100005];
namespace LCT {
int fa[100005], ch[100005][2], siz[100005], xsz[100005];
long long all[100005], ans[100005], sz2[100005], xans[100005];
bool isr(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
bool isc(int x) { return ch[fa[x]][1] == x; }
void upd(int x) {
  siz[x] = siz[ch[x][0]] + siz[ch[x][1]] + xsz[x];
  all[x] = all[ch[x][0]] + all[ch[x][1]] + 1ll * a[x] * xsz[x];
  ans[x] =
      ans[ch[x][0]] + ans[ch[x][1]] + xans[x] +
      a[x] * (1ll * xsz[x] * xsz[x] - sz2[x] + 2ll * xsz[x] * siz[ch[x][1]]) +
      2ll * all[ch[x][0]] * (siz[x] - siz[ch[x][0]]);
}
void rot(int x) {
  int y = fa[x], z = fa[y], c = isc(x);
  !isr(y) && (ch[z][isc(y)] = x);
  fa[ch[y][c] = ch[x][!c]] = y, fa[ch[x][!c] = y] = x, fa[x] = z;
  upd(y);
}
void splay(int x) {
  for (; !isr(x); rot(x))
    if (!isr(fa[x])) rot(isc(fa[x]) == isc(x) ? fa[x] : x);
  upd(x);
}
void access(int x) {
  for (int y = 0; x; x = fa[y = x]) {
    splay(x);
    xsz[x] += siz[ch[x][1]] - siz[y];
    sz2[x] += 1ll * siz[ch[x][1]] * siz[ch[x][1]] - 1ll * siz[y] * siz[y];
    xans[x] += ans[ch[x][1]] - ans[y];
    ch[x][1] = y, upd(x);
  }
}
void prefer(int x) { access(x), splay(x); }
void link(int x, int y) {
  splay(x), prefer(y);
  fa[x] = y, xsz[y] += siz[x], sz2[y] += 1ll * siz[x] * siz[x],
  xans[y] += ans[x], upd(y);
}
void cut(int x) { prefer(x), ch[x][0] = fa[ch[x][0]] = 0, upd(x); }
}  // namespace LCT
using namespace LCT;
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) scanf("%d", &F[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), xsz[i] = 1;
  for (int i = 2; i <= n; i++) link(i, F[i]);
  splay(1), printf("%.9f\n", 1. * ans[1] / n / n);
  scanf("%d", &m);
  char op[2];
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%s%d%d", op, &x, &y);
    if (op[0] == 'P') {
      if (prefer(y), splay(x), !isr(y)) swap(x, y);
      cut(x), link(x, F[x] = y);
      printf("%.9f\n", 1. * ans[y] / n / n);
    } else
      prefer(x), a[x] = y, upd(x), printf("%.9f\n", 1. * ans[x] / n / n);
  }
}
