#include <bits/stdc++.h>
using namespace std;
const int n_max = 300010, module = 1000000007;
list<int> G[n_max];
int T1[n_max * 2], T2[n_max * 2], Op[n_max], Ed[n_max], H[n_max];
int n, m;
int moded(long long x) {
  if (x < 0) x %= module, x += module;
  if (x >= module) x %= module;
  return x;
}
void upd(int t[], int x, int y) {
  while (x <= m) t[x] = moded(t[x] + y), x += x & -x;
}
int cal(int t[], int x) {
  int s = 0;
  while (x) s = moded(s + t[x]), x -= x & -x;
  return s;
}
void dfs(int x, int p) {
  H[x] = H[p] + 1;
  Op[x] = ++m;
  for (int y : G[x]) dfs(y, x);
  Ed[x] = ++m;
}
void enter() {
  cin >> n;
  for (int x = 2; x <= n; ++x) {
    int p;
    cin >> p;
    G[p].emplace_back(x);
  }
  dfs(1, 0);
}
void solve() {
  int q, t, x, v, k;
  cin >> q;
  while (q--) {
    cin >> t >> v;
    if (t == 2)
      cout << moded(cal(T1, Op[v]) - 1LL * H[v] * cal(T2, Op[v])) << "\n";
    else {
      cin >> x >> k;
      upd(T1, Op[v], moded(x + 1LL * H[v] * k));
      upd(T1, Ed[v], moded(-x - 1LL * H[v] * k));
      upd(T2, Op[v], k);
      upd(T2, Ed[v], -k);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  if (ifstream("test.inp")) cin.rdbuf((new ifstream("test.inp"))->rdbuf());
  enter();
  solve();
}

### Prompt

In cpp, your task is to solve 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_max = 300010, module = 1000000007;
list<int> G[n_max];
int T1[n_max * 2], T2[n_max * 2], Op[n_max], Ed[n_max], H[n_max];
int n, m;
int moded(long long x) {
  if (x < 0) x %= module, x += module;
  if (x >= module) x %= module;
  return x;
}
void upd(int t[], int x, int y) {
  while (x <= m) t[x] = moded(t[x] + y), x += x & -x;
}
int cal(int t[], int x) {
  int s = 0;
  while (x) s = moded(s + t[x]), x -= x & -x;
  return s;
}
void dfs(int x, int p) {
  H[x] = H[p] + 1;
  Op[x] = ++m;
  for (int y : G[x]) dfs(y, x);
  Ed[x] = ++m;
}
void enter() {
  cin >> n;
  for (int x = 2; x <= n; ++x) {
    int p;
    cin >> p;
    G[p].emplace_back(x);
  }
  dfs(1, 0);
}
void solve() {
  int q, t, x, v, k;
  cin >> q;
  while (q--) {
    cin >> t >> v;
    if (t == 2)
      cout << moded(cal(T1, Op[v]) - 1LL * H[v] * cal(T2, Op[v])) << "\n";
    else {
      cin >> x >> k;
      upd(T1, Op[v], moded(x + 1LL * H[v] * k));
      upd(T1, Ed[v], moded(-x - 1LL * H[v] * k));
      upd(T2, Op[v], k);
      upd(T2, Ed[v], -k);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  if (ifstream("test.inp")) cin.rdbuf((new ifstream("test.inp"))->rdbuf());
  enter();
  solve();
}

```