Problem: You are given a rooted tree on n vertices. The vertices are numbered from 1 to n; the root is the vertex number 1.

Each vertex has two integers associated with it: a_i and b_i. We denote the set of all ancestors of v (including v itself) by R(v). The awesomeness of a vertex v is defined as 

$$$\left| ∑_{w ∈ R(v)} a_w\right| ⋅ \left|∑_{w ∈ R(v)} b_w\right|,$$$

where |x| denotes the absolute value of x. 

Process q queries of one of the following forms: 

  * 1 v x — increase a_v by a positive integer x. 
  * 2 v — report the maximum awesomeness in the subtree of vertex v. 

Input

The first line contains two integers n and q (1 ≤ n ≤ 2⋅ 10^5, 1 ≤ q ≤ 10^5) — the number of vertices in the tree and the number of queries, respectively.

The second line contains n - 1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i means that there is an edge between vertices i and p_i.

The third line contains n integers a_1, a_2, ..., a_n (-5000 ≤ a_i ≤ 5000), the initial values of a_i for each vertex.

The fourth line contains n integers b_1, b_2, ..., b_n (-5000 ≤ b_i ≤ 5000), the values of b_i for each vertex.

Each of the next q lines describes a query. It has one of the following forms: 

  * 1 v x (1 ≤ v ≤ n, 1≤ x ≤ 5000). 
  * 2 v (1 ≤ v ≤ n). 

Output

For each query of the second type, print a single line with the maximum awesomeness in the respective subtree.

Example

Input

5 6
1 1 2 2
10 -3 -7 -3 -10
10 3 9 3 6
2 1
2 2
1 2 6
2 1
1 2 5
2 1


Output

100
91
169
240

Note

The initial awesomeness of the vertices is [100, 91, 57, 64, 57]. The most awesome vertex in the subtree of vertex 1 (the first query) is 1, and the most awesome vertex in the subtree of vertex 2 (the second query) is 2. 

After the first update (the third query), the awesomeness changes to [100, 169, 57, 160, 57] and thus the most awesome vertex in the whole tree (the fourth query) is now 2.

After the second update (the fifth query), the awesomeness becomes [100, 234, 57, 240, 152], hence the most awesome vertex (the sixth query) is now 4. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 200005;
int n, q, cnt, per, dfn_cnt, pos[Maxn], bel[Maxn], head[Maxn], siz[Maxn],
    dfn[Maxn], pnt[Maxn], top[Maxn];
long long a[Maxn], b[Maxn], A[Maxn], B[Maxn], shift[Maxn];
pair<long long, long long> sta[1105][2205];
struct edg {
  int nxt, to;
} edge[2 * Maxn];
void add(int x, int y) {
  edge[++cnt] = (edg){head[x], y};
  head[x] = cnt;
}
void dfs(int u) {
  dfn[u] = ++dfn_cnt;
  pos[dfn_cnt] = u;
  siz[u] = 1;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int to = edge[i].to;
    b[to] += b[u], a[to] += a[u], dfs(to), siz[u] += siz[to];
  }
}
double slope(pair<long long, long long> x, pair<long long, long long> y) {
  return (x.second - y.second) / (double)(y.first - x.first);
}
vector<pair<long long, long long> > tmp;
void rebuild(int x) {
  int lt = (x - 1) * per + 1, rt = min(n, x * per);
  for (int i = lt; i <= rt; i++) A[i] += shift[x];
  shift[x] = 0;
  tmp.clear();
  for (int i = lt; i <= rt; i++)
    tmp.push_back(make_pair(B[i], A[i] * B[i])),
        tmp.push_back(make_pair(-B[i], -A[i] * B[i]));
  sort(tmp.begin(), tmp.end());
  pnt[x] = 1, top[x] = 0;
  for (auto now : tmp) {
    while (top[x] && sta[x][top[x]].first == now.first) top[x]--;
    while (top[x] > 1 && slope(sta[x][top[x] - 1], sta[x][top[x]]) >=
                             slope(sta[x][top[x]], now))
      top[x]--;
    sta[x][++top[x]] = now;
  }
}
void modify(int lt, int rt, int val) {
  if (bel[lt] == bel[rt]) {
    for (int i = lt; i <= rt; i++) A[i] += val;
    rebuild(bel[lt]);
  } else {
    for (int i = lt; i <= bel[lt] * per; i++) A[i] += val;
    rebuild(bel[lt]);
    for (int i = (bel[rt] - 1) * per + 1; i <= rt; i++) A[i] += val;
    rebuild(bel[rt]);
    for (int i = bel[lt] + 1; i <= bel[rt] - 1; i++) shift[i] += val;
  }
}
long long ask(int lt, int rt) {
  long long ans = 0;
  if (bel[lt] == bel[rt]) {
    for (int i = lt; i <= rt; i++)
      ans = max(ans, abs(A[i] + shift[bel[lt]]) * B[i]);
    return ans;
  }
  for (int i = lt; i <= bel[lt] * per; i++)
    ans = max(ans, abs(A[i] + shift[bel[lt]]) * B[i]);
  for (int i = (bel[rt] - 1) * per + 1; i <= rt; i++)
    ans = max(ans, abs(A[i] + shift[bel[rt]]) * B[i]);
  for (int i = bel[lt] + 1; i <= bel[rt] - 1; i++) {
    while (pnt[i] < top[i] &&
           sta[i][pnt[i]].first * shift[i] + sta[i][pnt[i]].second <=
               sta[i][pnt[i] + 1].first * shift[i] + sta[i][pnt[i] + 1].second)
      pnt[i]++;
    ans = max(ans, sta[i][pnt[i]].first * shift[i] + sta[i][pnt[i]].second);
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    add(x, i);
  }
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%lld", &b[i]);
  dfs(1);
  for (int i = 1; i <= n; i++) A[dfn[i]] = a[i], B[dfn[i]] = abs(b[i]);
  int bloc = max(1.0, sqrt(n * 6));
  per = n / bloc + 1;
  for (int i = 1; i <= n; i++) bel[i] = (i - 1) / per + 1;
  for (int i = 1; i <= bloc; i++) rebuild(i);
  for (int i = 1; i <= q; i++) {
    int opt;
    scanf("%d", &opt);
    if (opt == 1) {
      int v, x;
      scanf("%d%d", &v, &x);
      modify(dfn[v], dfn[v] + siz[v] - 1, x);
    } else {
      int v;
      scanf("%d", &v);
      printf("%lld\n", ask(dfn[v], dfn[v] + siz[v] - 1));
    }
  }
  return 0;
}