#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10, mod = 1e9 + 7;
int h[maxn], n;
vector<int> adj[maxn], vec;
int tin[maxn], tout[maxn], timer = -1;
void dfs(int u = 1) {
  vec.push_back(h[u]);
  tin[u] = ++timer;
  for (int v : adj[u]) {
    h[v] = h[u] + 1;
    dfs(v);
  }
  tout[u] = timer;
}
pair<int, int> seg[maxn << 2];
void add(int v, int x, int y) {
  (seg[v].first += x) %= mod;
  (seg[v].second += y) %= mod;
}
void interval_add(int l, int r, int x, int y, int v = 1, int tl = 0,
                  int tr = n - 1) {
  if (l > r) return;
  if (tl == l && tr == r) return add(v, x, y);
  int tm = tl + tr >> 1;
  interval_add(l, min(r, tm), x, y, v << 1, tl, tm);
  interval_add(max(l, tm + 1), r, x, y, v << 1 | 1, tm + 1, tr);
}
int get(int idx, int cur = 0, int cnt = 0, int v = 1, int tl = 0,
        int tr = n - 1) {
  (cur += seg[v].first) %= mod;
  (cnt += seg[v].second) %= mod;
  if (tl == tr) {
    return (cur + 1ll * cnt * vec[tl] % mod) % mod;
  }
  int tm = tl + tr >> 1;
  return idx <= tm ? get(idx, cur, cnt, v << 1, tl, tm)
                   : get(idx, cur, cnt, v << 1 | 1, tm + 1, tr);
}
void add_val(int v, int x, int k) {
  interval_add(tin[v], tout[v], (x + 1ll * h[v] * k % mod) % mod, mod - k);
}
int get_val(int v) { return get(tin[v]); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 2, p; i <= n; i++) {
    cin >> p;
    adj[p].push_back(i);
  }
  dfs();
  int q;
  cin >> q;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int v, x, k;
      cin >> v >> x >> k;
      add_val(v, x, k);
    } else {
      int v;
      cin >> v;
      cout << get_val(v) << '\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 maxn = 3e5 + 10, mod = 1e9 + 7;
int h[maxn], n;
vector<int> adj[maxn], vec;
int tin[maxn], tout[maxn], timer = -1;
void dfs(int u = 1) {
  vec.push_back(h[u]);
  tin[u] = ++timer;
  for (int v : adj[u]) {
    h[v] = h[u] + 1;
    dfs(v);
  }
  tout[u] = timer;
}
pair<int, int> seg[maxn << 2];
void add(int v, int x, int y) {
  (seg[v].first += x) %= mod;
  (seg[v].second += y) %= mod;
}
void interval_add(int l, int r, int x, int y, int v = 1, int tl = 0,
                  int tr = n - 1) {
  if (l > r) return;
  if (tl == l && tr == r) return add(v, x, y);
  int tm = tl + tr >> 1;
  interval_add(l, min(r, tm), x, y, v << 1, tl, tm);
  interval_add(max(l, tm + 1), r, x, y, v << 1 | 1, tm + 1, tr);
}
int get(int idx, int cur = 0, int cnt = 0, int v = 1, int tl = 0,
        int tr = n - 1) {
  (cur += seg[v].first) %= mod;
  (cnt += seg[v].second) %= mod;
  if (tl == tr) {
    return (cur + 1ll * cnt * vec[tl] % mod) % mod;
  }
  int tm = tl + tr >> 1;
  return idx <= tm ? get(idx, cur, cnt, v << 1, tl, tm)
                   : get(idx, cur, cnt, v << 1 | 1, tm + 1, tr);
}
void add_val(int v, int x, int k) {
  interval_add(tin[v], tout[v], (x + 1ll * h[v] * k % mod) % mod, mod - k);
}
int get_val(int v) { return get(tin[v]); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 2, p; i <= n; i++) {
    cin >> p;
    adj[p].push_back(i);
  }
  dfs();
  int q;
  cin >> q;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int v, x, k;
      cin >> v >> x >> k;
      add_val(v, x, k);
    } else {
      int v;
      cin >> v;
      cout << get_val(v) << '\n';
    }
  }
  return 0;
}

```