#include <bits/stdc++.h>
using namespace std;
const int N = 300 * 1000 + 10;
const long long MOD = 1000 * 1000 * 1000 + 7;
vector<int> graph[N];
int h[N], st[N], cnt, en[N];
pair<long long, long long> t[N * 4];
void dfs(int v) {
  st[v] = cnt++;
  for (int i = 0; i < (int)graph[v].size(); i++) {
    int u = graph[v][i];
    h[u] = h[v] + 1;
    dfs(u);
  }
  en[v] = cnt;
}
void upd(int l, int r, pair<long long, long long> p, int v, int s, int e) {
  if (l <= s && e <= r) {
    t[v].first += p.first;
    t[v].first %= MOD;
    t[v].second += p.second;
    t[v].second %= MOD;
    return;
  }
  if (e - s < 2 || r <= s || l >= e) return;
  int mid = (s + e) / 2;
  upd(l, r, p, v * 2, s, mid);
  upd(l, r, p, v * 2 + 1, mid, e);
}
pair<long long, long long> get(int p, int v, int s, int e) {
  if (e - s < 2) return t[v];
  int mid = (s + e) / 2;
  pair<long long, long long> ans;
  if (p < mid)
    ans = get(p, 2 * v, s, mid);
  else
    ans = get(p, 2 * v + 1, mid, e);
  ans.first += t[v].first;
  ans.first %= MOD;
  ans.second += t[v].second;
  ans.second %= MOD;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x;
    cin >> x;
    x--;
    graph[x].push_back(i);
  }
  dfs(0);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, t;
    cin >> t >> v;
    v--;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      upd(st[v], en[v],
          pair<long long, long long>((x + h[v] * k % MOD) % MOD, k), 1, 0, n);
    } else {
      pair<long long, long long> ans = get(st[v], 1, 0, n);
      cout << (ans.first + MOD - h[v] * ans.second % MOD) % MOD << endl;
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 300 * 1000 + 10;
const long long MOD = 1000 * 1000 * 1000 + 7;
vector<int> graph[N];
int h[N], st[N], cnt, en[N];
pair<long long, long long> t[N * 4];
void dfs(int v) {
  st[v] = cnt++;
  for (int i = 0; i < (int)graph[v].size(); i++) {
    int u = graph[v][i];
    h[u] = h[v] + 1;
    dfs(u);
  }
  en[v] = cnt;
}
void upd(int l, int r, pair<long long, long long> p, int v, int s, int e) {
  if (l <= s && e <= r) {
    t[v].first += p.first;
    t[v].first %= MOD;
    t[v].second += p.second;
    t[v].second %= MOD;
    return;
  }
  if (e - s < 2 || r <= s || l >= e) return;
  int mid = (s + e) / 2;
  upd(l, r, p, v * 2, s, mid);
  upd(l, r, p, v * 2 + 1, mid, e);
}
pair<long long, long long> get(int p, int v, int s, int e) {
  if (e - s < 2) return t[v];
  int mid = (s + e) / 2;
  pair<long long, long long> ans;
  if (p < mid)
    ans = get(p, 2 * v, s, mid);
  else
    ans = get(p, 2 * v + 1, mid, e);
  ans.first += t[v].first;
  ans.first %= MOD;
  ans.second += t[v].second;
  ans.second %= MOD;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x;
    cin >> x;
    x--;
    graph[x].push_back(i);
  }
  dfs(0);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, t;
    cin >> t >> v;
    v--;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      upd(st[v], en[v],
          pair<long long, long long>((x + h[v] * k % MOD) % MOD, k), 1, 0, n);
    } else {
      pair<long long, long long> ans = get(st[v], 1, 0, n);
      cout << (ans.first + MOD - h[v] * ans.second % MOD) % MOD << endl;
    }
  }
  return 0;
}

```