#include <bits/stdc++.h>
using namespace std;
pair<int, int> add(pair<int, int> a, pair<int, int> b) {
  return {(a.first + b.first) % 1000000007, (a.second + b.second) % 1000000007};
}
pair<int, int> aint[1200010];
int l, r, poz;
pair<int, int> f;
void update(int nod, int st, int dr) {
  if (st > r || dr < l) return;
  if (l <= st && r >= dr) {
    aint[nod] = add(aint[nod], f);
    return;
  }
  update(2 * nod, st, (st + dr) / 2);
  update(2 * nod + 1, (st + dr) / 2 + 1, dr);
}
pair<int, int> query(int nod, int st, int dr) {
  if (st > poz || dr < poz) return {0, 0};
  if (st == dr) return aint[nod];
  return add(add(aint[nod], query(2 * nod, st, (st + dr) / 2)),
             query(2 * nod + 1, (st + dr) / 2 + 1, dr));
}
vector<int> adia[400010];
int firstap[400010], lastap[400010];
int h[400010];
int cnt;
void dfs(int nod, int tata) {
  h[nod] = 1 + h[tata];
  firstap[nod] = ++cnt;
  for (auto i : adia[nod])
    if (i != tata) dfs(i, nod);
  lastap[nod] = cnt;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n, a, b;
  cin >> n;
  for (int i(2); i <= n; i++) {
    cin >> a;
    adia[a].push_back(i);
  }
  dfs(1, 0);
  int q, v, x, k, t;
  cin >> q;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> v >> x >> k;
      f = {(x + 1ll * h[v] * k) % 1000000007, -k};
      l = firstap[v], r = lastap[v];
      update(1, 1, cnt);
    } else {
      cin >> v;
      poz = firstap[v];
      pair<int, int> z = query(1, 1, cnt);
      cout << ((z.first + 1ll * h[v] * z.second) % 1000000007 + 1000000007) %
                  1000000007
           << '\n';
    }
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

You are given a rooted tree consisting of n vertices numbered from 1 to n. The root of the tree is a vertex number 1.

Initially all vertices contain number 0. Then come q queries, each query has one of the two types:

  * The format of the query: 1 v x k. In response to the query, you need to add to the number at vertex v number x; to the numbers at the descendants of vertex v at distance 1, add x - k; and so on, to the numbers written in the descendants of vertex v at distance i, you need to add x - (i·k). The distance between two vertices is the number of edges in the shortest path between these vertices. 
  * The format of the query: 2 v. In reply to the query you should print the number written in vertex v modulo 1000000007 (109 + 7). 



Process the queries given in the input.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the number of vertices in the tree. The second line contains n - 1 integers p2, p3, ... pn (1 ≤ pi < i), where pi is the number of the vertex that is the parent of vertex i in the tree.

The third line contains integer q (1 ≤ q ≤ 3·105) — the number of queries. Next q lines contain the queries, one per line. The first number in the line is type. It represents the type of the query. If type = 1, then next follow space-separated integers v, x, k (1 ≤ v ≤ n; 0 ≤ x < 109 + 7; 0 ≤ k < 109 + 7). If type = 2, then next follows integer v (1 ≤ v ≤ n) — the vertex where you need to find the value of the number.

Output

For each query of the second type print on a single line the number written in the vertex from the query. Print the number modulo 1000000007 (109 + 7).

Examples

Input

3
1 1
3
1 1 2 1
2 1
2 2


Output

2
1

Note

You can read about a rooted tree here: http://en.wikipedia.org/wiki/Tree_(graph_theory).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
pair<int, int> add(pair<int, int> a, pair<int, int> b) {
  return {(a.first + b.first) % 1000000007, (a.second + b.second) % 1000000007};
}
pair<int, int> aint[1200010];
int l, r, poz;
pair<int, int> f;
void update(int nod, int st, int dr) {
  if (st > r || dr < l) return;
  if (l <= st && r >= dr) {
    aint[nod] = add(aint[nod], f);
    return;
  }
  update(2 * nod, st, (st + dr) / 2);
  update(2 * nod + 1, (st + dr) / 2 + 1, dr);
}
pair<int, int> query(int nod, int st, int dr) {
  if (st > poz || dr < poz) return {0, 0};
  if (st == dr) return aint[nod];
  return add(add(aint[nod], query(2 * nod, st, (st + dr) / 2)),
             query(2 * nod + 1, (st + dr) / 2 + 1, dr));
}
vector<int> adia[400010];
int firstap[400010], lastap[400010];
int h[400010];
int cnt;
void dfs(int nod, int tata) {
  h[nod] = 1 + h[tata];
  firstap[nod] = ++cnt;
  for (auto i : adia[nod])
    if (i != tata) dfs(i, nod);
  lastap[nod] = cnt;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int n, a, b;
  cin >> n;
  for (int i(2); i <= n; i++) {
    cin >> a;
    adia[a].push_back(i);
  }
  dfs(1, 0);
  int q, v, x, k, t;
  cin >> q;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> v >> x >> k;
      f = {(x + 1ll * h[v] * k) % 1000000007, -k};
      l = firstap[v], r = lastap[v];
      update(1, 1, cnt);
    } else {
      cin >> v;
      poz = firstap[v];
      pair<int, int> z = query(1, 1, cnt);
      cout << ((z.first + 1ll * h[v] * z.second) % 1000000007 + 1000000007) %
                  1000000007
           << '\n';
    }
  }
  return 0;
}

```