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 bool CHECK_OVERFLOW64 = true;
const long long LL_INF = (long long)2e18 + 5;
struct point {
  long long x, y;
  point() : x(0), y(0) {}
  point(long long _x, long long _y) : x(_x), y(_y) {}
  point operator-(const point &other) const {
    return point(x - other.x, y - other.y);
  }
};
long long cross(const point &a, const point &b) {
  return a.x * b.y - b.x * a.y;
}
int cross_sign(const point &a, const point &b) {
  if (CHECK_OVERFLOW64) {
    double double_value = (double)a.x * b.y - (double)b.x * a.y;
    if (abs(double_value) > 1e12) return double_value > 0 ? +1 : -1;
  }
  uint64_t uint64_value = (uint64_t)a.x * b.y - (uint64_t)b.x * a.y;
  if (uint64_value == 0) return 0;
  return uint64_value >> 63 ? -1 : +1;
}
bool left_turn(const point &a, const point &b, const point &c) {
  return cross_sign(b - a, c - b) > 0;
}
struct monotonic_dp_hull {
  vector<point> points;
  int front = 0, end = 0;
  void init(int capacity) { points.resize(capacity); }
  int size() const { return end - front; }
  void clear() { front = end = 0; }
  void insert(const point &p) {
    assert(size() == 0 || p.x >= points[end - 1].x);
    if (size() > 0 && p.x == points[end - 1].x) {
      if (p.y <= points[end - 1].y) return;
      points.pop_back();
    }
    while (size() >= 2 && left_turn(points[end - 2], points[end - 1], p)) end--;
    points[end++] = p;
  }
  void insert(long long a, long long b) { insert(point(a, b)); }
  long long query(long long x) {
    while (size() >= 2 && x * (points[front + 1].x - points[front].x) >=
                              points[front].y - points[front + 1].y)
      front++;
    return points[front].x * x + points[front].y;
  }
};
struct vertex {
  long long A = 0, B = 0;
  int index;
  bool operator<(const vertex &other) const { return B < other.B; }
};
int N, Q;
int BUCKET_SIZE;
vector<int> parent;
vector<vector<int>> children;
vector<vertex> vertices;
vector<vertex> sums, sorted_sums;
vector<int> tour_start, tour_end;
void dfs(int node) {
  static int tour;
  if (parent[node] < 0) tour = 0;
  tour_start[node] = tour++;
  sums[tour_start[node]].A =
      vertices[node].A +
      (parent[node] < 0 ? 0 : sums[tour_start[parent[node]]].A);
  sums[tour_start[node]].B =
      vertices[node].B +
      (parent[node] < 0 ? 0 : sums[tour_start[parent[node]]].B);
  for (int child : children[node]) dfs(child);
  tour_end[node] = tour;
}
vector<monotonic_dp_hull> hulls;
vector<long long> bucket_add;
int get_bucket_start(int index) { return index - index % BUCKET_SIZE; }
int get_bucket_end_from_start(int start) { return min(start + BUCKET_SIZE, N); }
void build_bucket(int start, int end) {
  hulls[start].init(2 * (end - start));
  hulls[start].clear();
  for (int i = end - 1; i >= start; i--)
    hulls[start].insert(-sorted_sums[i].B,
                        -sorted_sums[i].A * sorted_sums[i].B);
  for (int i = start; i < end; i++)
    hulls[start].insert(sorted_sums[i].B, sorted_sums[i].A * sorted_sums[i].B);
}
void build() {
  BUCKET_SIZE = 0.5 * sqrt(N) + 1;
  sorted_sums = sums;
  hulls.resize(N);
  bucket_add.assign(N, 0);
  for (int start = 0, end = 0; start < N; start = end) {
    end = get_bucket_end_from_start(start);
    sort(sorted_sums.begin() + start, sorted_sums.begin() + end);
    build_bucket(start, end);
  }
}
void update_bucket(int start, int x) { bucket_add[start] += x; }
long long query_bucket(int start) {
  return hulls[start].query(bucket_add[start]);
}
void update_partial_bucket(int bucket_start, int bucket_end, int start, int end,
                           int x) {
  for (int i = start; i < end; i++) sums[i].A += x;
  for (int i = bucket_start; i < bucket_end; i++)
    if (start <= sorted_sums[i].index && sorted_sums[i].index < end)
      sorted_sums[i].A += x;
  build_bucket(bucket_start, bucket_end);
}
long long query_partial_bucket(int bucket_start, int start, int end) {
  long long most = 0;
  for (int i = start; i < end; i++)
    most = max(most, abs(sums[i].A + bucket_add[bucket_start]) * sums[i].B);
  return most;
}
void update(int start, int end, int x) {
  int bucket_start = get_bucket_start(start);
  int bucket_end = get_bucket_end_from_start(bucket_start);
  if (bucket_start < start) {
    update_partial_bucket(bucket_start, bucket_end, start, min(end, bucket_end),
                          x);
    start = bucket_end;
  }
  while (start < end && get_bucket_end_from_start(start) <= end) {
    update_bucket(start, x);
    start = get_bucket_end_from_start(start);
  }
  if (start < end)
    update_partial_bucket(start, get_bucket_end_from_start(start), start, end,
                          x);
}
long long query(int start, int end) {
  long long best = 0;
  int bucket_start = get_bucket_start(start);
  int bucket_end = get_bucket_end_from_start(bucket_start);
  if (bucket_start < start) {
    best = query_partial_bucket(bucket_start, start, min(end, bucket_end));
    start = bucket_end;
  }
  while (start < end && get_bucket_end_from_start(start) <= end) {
    best = max(best, query_bucket(start));
    start = get_bucket_end_from_start(start);
  }
  if (start < end) best = max(best, query_partial_bucket(start, start, end));
  return best;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> Q;
  parent.assign(N, -1);
  children.assign(N, {});
  vertices.resize(N);
  sums.resize(N);
  tour_start.resize(N);
  tour_end.resize(N);
  for (int i = 1; i < N; i++) {
    cin >> parent[i];
    parent[i]--;
    children[parent[i]].push_back(i);
  }
  for (int i = 0; i < N; i++) cin >> vertices[i].A;
  for (int i = 0; i < N; i++) cin >> vertices[i].B;
  dfs(0);
  for (int i = 0; i < N; i++) {
    vertices[i].index = i;
    sums[i].index = i;
    sums[i].B = abs(sums[i].B);
  }
  build();
  for (int q = 0; q < Q; q++) {
    int type, v, x;
    cin >> type >> v;
    v--;
    if (type == 1) {
      cin >> x;
      update(tour_start[v], tour_end[v], x);
    } else if (type == 2) {
      cout << query(tour_start[v], tour_end[v]) << '\n';
    } else {
      assert(false);
    }
  }
}