#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, mod = 1e9 + 7;
int st[N], ed[N], n, q, h[N], f[N], g[N];
vector<int> adj[N];
void dfs(int v) {
  static int cnt = 0;
  st[v] = ++cnt;
  for (int i = 0; i < adj[v].size(); i++) dfs(adj[v][i]);
  ed[v] = cnt;
}
void add(int *f, int m, int x) {
  while (m) {
    f[m] = (f[m] + x) % mod;
    m -= m & -m;
  }
}
int query(int *f, int m) {
  int res = 0;
  while (m < N) {
    res = (res + f[m]) % mod;
    m += m & -m;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int p;
    cin >> p;
    p--;
    adj[p].push_back(i);
    h[i] = h[p] + 1;
  }
  dfs(0);
  cin >> q;
  while (q--) {
    int t, v, x, k;
    cin >> t;
    if (t == 1) {
      cin >> v >> x >> k;
      v--;
      add(f, ed[v], (x + 1ll * k * h[v]) % mod);
      add(f, st[v] - 1, -(x + 1ll * k * h[v]) % mod);
      add(g, ed[v], k);
      add(g, st[v] - 1, -k);
    } else {
      cin >> v;
      v--;
      cout << ((query(f, st[v]) - 1ll * h[v] * query(g, st[v])) % mod + mod) %
                  mod
           << '\n';
    }
  }
  return 0;
}

### Prompt

Generate 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;
const int N = 3e5 + 5, mod = 1e9 + 7;
int st[N], ed[N], n, q, h[N], f[N], g[N];
vector<int> adj[N];
void dfs(int v) {
  static int cnt = 0;
  st[v] = ++cnt;
  for (int i = 0; i < adj[v].size(); i++) dfs(adj[v][i]);
  ed[v] = cnt;
}
void add(int *f, int m, int x) {
  while (m) {
    f[m] = (f[m] + x) % mod;
    m -= m & -m;
  }
}
int query(int *f, int m) {
  int res = 0;
  while (m < N) {
    res = (res + f[m]) % mod;
    m += m & -m;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int p;
    cin >> p;
    p--;
    adj[p].push_back(i);
    h[i] = h[p] + 1;
  }
  dfs(0);
  cin >> q;
  while (q--) {
    int t, v, x, k;
    cin >> t;
    if (t == 1) {
      cin >> v >> x >> k;
      v--;
      add(f, ed[v], (x + 1ll * k * h[v]) % mod);
      add(f, st[v] - 1, -(x + 1ll * k * h[v]) % mod);
      add(g, ed[v], k);
      add(g, st[v] - 1, -k);
    } else {
      cin >> v;
      v--;
      cout << ((query(f, st[v]) - 1ll * h[v] * query(g, st[v])) % mod + mod) %
                  mod
           << '\n';
    }
  }
  return 0;
}

```