#include <bits/stdc++.h>
using namespace std;
const int MN = 3e5 + 7;
const int Mo = 1e9 + 7;
int n, m, tot;
int d[MN], l[MN], r[MN], a[MN], b[MN];
vector<int> f[MN];
void dfs(int s) {
  l[s] = ++tot;
  for (int i = 0; i < f[s].size(); i++) {
    d[f[s][i]] = d[s] + 1;
    dfs(f[s][i]);
  }
  r[s] = tot;
}
void add(int i, int u, int v) {
  for (; i <= n; i += i & -i) {
    a[i] = (a[i] + u) % Mo;
    b[i] = (b[i] + v) % Mo;
  }
}
int get(int i, long long d) {
  long long ans = 0;
  for (; i; i -= i & -i) ans = (ans + a[i] + d * b[i]) % Mo;
  return (ans + Mo) % Mo;
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  for (i = 2; i <= n; i++) {
    scanf("%d", &k);
    f[k].push_back(i);
  }
  dfs(1);
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &k);
    if (k == 1) {
      int x, v, w;
      scanf("%d%d%d", &x, &v, &w);
      long long tmp = (v + (long long)d[x] * w);
      add(l[x], tmp % Mo, -w);
      add(r[x] + 1, (-tmp) % Mo, w);
    } else {
      scanf("%d", &i);
      printf("%d\n", get(l[i], d[i]));
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 MN = 3e5 + 7;
const int Mo = 1e9 + 7;
int n, m, tot;
int d[MN], l[MN], r[MN], a[MN], b[MN];
vector<int> f[MN];
void dfs(int s) {
  l[s] = ++tot;
  for (int i = 0; i < f[s].size(); i++) {
    d[f[s][i]] = d[s] + 1;
    dfs(f[s][i]);
  }
  r[s] = tot;
}
void add(int i, int u, int v) {
  for (; i <= n; i += i & -i) {
    a[i] = (a[i] + u) % Mo;
    b[i] = (b[i] + v) % Mo;
  }
}
int get(int i, long long d) {
  long long ans = 0;
  for (; i; i -= i & -i) ans = (ans + a[i] + d * b[i]) % Mo;
  return (ans + Mo) % Mo;
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  for (i = 2; i <= n; i++) {
    scanf("%d", &k);
    f[k].push_back(i);
  }
  dfs(1);
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &k);
    if (k == 1) {
      int x, v, w;
      scanf("%d%d%d", &x, &v, &w);
      long long tmp = (v + (long long)d[x] * w);
      add(l[x], tmp % Mo, -w);
      add(r[x] + 1, (-tmp) % Mo, w);
    } else {
      scanf("%d", &i);
      printf("%d\n", get(l[i], d[i]));
    }
  }
  return 0;
}

```