#include <bits/stdc++.h>
using namespace std;
long long int BIT[2][300005];
void update(int cs, int indx, long long int val) {
  while (indx < 300005) {
    BIT[cs][indx] += val;
    BIT[cs][indx] %= 1000000007;
    indx += (indx & -indx);
  }
}
int sum(int cs, int indx) {
  int ans = 0;
  while (indx != 0) {
    ans += BIT[cs][indx];
    ans %= 1000000007;
    indx -= (indx & -indx);
  }
  return ans;
}
int Time;
int P[300005];
int T[300005];
int D[300005];
int st[300005];
int en[300005];
vector<int> adj[300005];
void dfs(int u, int d) {
  st[u] = ++Time;
  D[u] = d;
  for (int i = 0; i < adj[u].size(); i++) dfs(adj[u][i], d + 1);
  en[u] = Time;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &P[i]);
    adj[P[i]].push_back(i);
  }
  dfs(1, 0);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int cmd;
    scanf("%d", &cmd);
    if (cmd == 1) {
      int v;
      long long int x, k;
      scanf("%d %I64d %I64d", &v, &x, &k);
      long long int add = (x + D[v] * k) % 1000000007;
      update(0, st[v], add);
      update(0, en[v] + 1, 1000000007 - add);
      update(1, st[v], k);
      update(1, en[v] + 1, -k);
    } else {
      int v;
      scanf("%d", &v);
      long long int Ans = sum(0, st[v]);
      long long int Count = sum(1, st[v]);
      Ans -= (Count * D[v]) % 1000000007;
      Ans = (Ans + 1000000007) % 1000000007;
      printf("%I64d\n", Ans);
    }
  }
}

### 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;
long long int BIT[2][300005];
void update(int cs, int indx, long long int val) {
  while (indx < 300005) {
    BIT[cs][indx] += val;
    BIT[cs][indx] %= 1000000007;
    indx += (indx & -indx);
  }
}
int sum(int cs, int indx) {
  int ans = 0;
  while (indx != 0) {
    ans += BIT[cs][indx];
    ans %= 1000000007;
    indx -= (indx & -indx);
  }
  return ans;
}
int Time;
int P[300005];
int T[300005];
int D[300005];
int st[300005];
int en[300005];
vector<int> adj[300005];
void dfs(int u, int d) {
  st[u] = ++Time;
  D[u] = d;
  for (int i = 0; i < adj[u].size(); i++) dfs(adj[u][i], d + 1);
  en[u] = Time;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &P[i]);
    adj[P[i]].push_back(i);
  }
  dfs(1, 0);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int cmd;
    scanf("%d", &cmd);
    if (cmd == 1) {
      int v;
      long long int x, k;
      scanf("%d %I64d %I64d", &v, &x, &k);
      long long int add = (x + D[v] * k) % 1000000007;
      update(0, st[v], add);
      update(0, en[v] + 1, 1000000007 - add);
      update(1, st[v], k);
      update(1, en[v] + 1, -k);
    } else {
      int v;
      scanf("%d", &v);
      long long int Ans = sum(0, st[v]);
      long long int Count = sum(1, st[v]);
      Ans -= (Count * D[v]) % 1000000007;
      Ans = (Ans + 1000000007) % 1000000007;
      printf("%I64d\n", Ans);
    }
  }
}

```