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 + 5;
int n, Q, a[N], fa[N];
inline int read() {
  int s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') f = ch == '-' ? -1 : f, ch = getchar();
  while (ch >= '0' && ch <= '9')
    s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
  return s * f;
}
struct LCT {
  int f[N], ch[N][2];
  long long sz[N], sum[N], ans[N], all[N], ad[N], de[N];
  inline bool nrt(int x) { return ch[f[x]][0] == x || ch[f[x]][1] == x; }
  inline void upd(int x) {
    sum[x] = sum[ch[x][0]] + sum[ch[x][1]] + sz[x];
    all[x] = all[ch[x][0]] + all[ch[x][1]] + a[x] * sz[x];
    ans[x] = ans[ch[x][0]] + ans[ch[x][1]] + ad[x] +
             a[x] * (sz[x] * sz[x] - de[x]) + 2 * a[x] * sz[x] * sum[ch[x][1]] +
             2 * all[ch[x][0]] * (sum[x] - sum[ch[x][0]]);
  }
  inline void rotate(int x) {
    int y = f[x], z = f[y], k = ch[y][1] == x, w = ch[x][!k];
    if (nrt(y)) ch[z][ch[z][1] == y] = x;
    ch[x][!k] = y;
    if (w) f[w] = y;
    f[y] = x;
    f[x] = z;
    ch[y][k] = w;
    upd(y), upd(x);
  }
  inline void splay(int x) {
    int y = x, z = 0;
    while (nrt(x)) {
      y = f[x], z = f[y];
      if (nrt(y)) rotate((ch[y][0] == x) ^ (ch[z][0] == y) ? x : y);
      rotate(x);
    }
    upd(x);
  }
  inline void access(int x) {
    for (int y = 0; x; x = f[y = x]) {
      splay(x);
      sz[x] += sum[ch[x][1]];
      de[x] += sum[ch[x][1]] * sum[ch[x][1]];
      ad[x] += ans[ch[x][1]];
      ch[x][1] = y;
      sz[x] -= sum[y];
      de[x] -= sum[y] * sum[y];
      ad[x] -= ans[y];
      upd(x);
    }
  }
  inline void link(int x, int y) {
    if (!x || !y) return;
    access(y);
    splay(y);
    access(x);
    splay(x);
    f[y] = x;
    ad[x] += ans[y];
    sz[x] += sum[y];
    de[x] += sum[y] * sum[y];
    upd(x);
    access(x);
    splay(x);
  }
  inline void cut(int x, int y) {
    access(x);
    splay(x);
    splay(y);
    sz[x] -= sum[y];
    de[x] -= sum[y] * sum[y];
    ad[x] -= ans[y];
    f[y] = ch[x][1] = 0;
    upd(x);
  }
  inline void init(int x) {
    sum[x] = sz[x] = 1;
    all[x] = ans[x] = a[x];
  }
  inline bool check(int x, int y) {
    access(y);
    splay(y);
    splay(x);
    return nrt(y);
  }
} lct;
int main() {
  n = read();
  for (int i = 2; i <= n; ++i) fa[i] = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i) lct.init(i), lct.upd(i);
  for (int i = 2; i <= n; ++i) lct.link(fa[i], i);
  lct.access(1);
  lct.splay(1);
  printf("%.10lf\n", 1.0 * lct.ans[1] / n / n);
  Q = read();
  char opt[3];
  for (int i = 1; i <= Q; ++i) {
    scanf("%s", opt + 1);
    int x = read(), y = read();
    if (opt[1] == 'P') {
      if (fa[x] == y || fa[y] == x) {
        printf("%.10lf\n", 1.0 * lct.ans[1] / n / n);
        continue;
      }
      if (lct.check(x, y)) swap(x, y);
      lct.cut(fa[x], x);
      lct.link(y, x);
      fa[x] = y;
      lct.access(1);
      lct.splay(1);
    }
    if (opt[1] == 'V') {
      lct.access(x);
      lct.splay(x);
      a[x] = y;
      lct.upd(x);
      lct.access(1);
      lct.splay(1);
    }
    printf("%.10lf\n", 1.0 * lct.ans[1] / n / n);
  }
}
