#include <bits/stdc++.h>
using namespace std;
const int mo = 1e9 + 7;
int n;
int l[300005], r[300005], cnt;
vector<int> e[300005];
long long a[300005], b[300005], d[300005];
void add(int x, long long y1, long long y2) {
  for (int i = x; i <= n; i += (i & -i)) {
    a[i] = (a[i] + y1) % mo;
    b[i] = (b[i] + y2) % mo;
  }
}
long long get(int x) {
  long long ret = 0;
  for (int i = x; i > 0; i -= (i & -i)) {
    ret = (ret + a[i]) % mo;
    ret = (ret + b[i] * d[x] % mo) % mo;
  }
  return ret;
}
void dfs(int k) {
  l[k] = ++cnt;
  for (int i = 0; i < e[k].size(); ++i) {
    int u = e[k][i];
    d[cnt + 1] = d[l[k]] + 1;
    dfs(u);
  }
  r[k] = cnt;
}
int main() {
  scanf("%d", &n);
  int t, v, x, k;
  for (int i = 2; i <= n; ++i) {
    scanf("%d", &x);
    e[x].push_back(i);
  }
  d[1] = 1;
  dfs(1);
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &v, &x, &k);
      add(l[v], (d[l[v]] * k % mo + x) % mo, -k);
      add(r[v] + 1, -(d[l[v]] * k % mo + x) % mo, k);
    } else {
      scanf("%d", &v);
      int ans = get(l[v]);
      if (ans < 0) ans += mo;
      printf("%d\n", ans);
    }
  }
  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 mo = 1e9 + 7;
int n;
int l[300005], r[300005], cnt;
vector<int> e[300005];
long long a[300005], b[300005], d[300005];
void add(int x, long long y1, long long y2) {
  for (int i = x; i <= n; i += (i & -i)) {
    a[i] = (a[i] + y1) % mo;
    b[i] = (b[i] + y2) % mo;
  }
}
long long get(int x) {
  long long ret = 0;
  for (int i = x; i > 0; i -= (i & -i)) {
    ret = (ret + a[i]) % mo;
    ret = (ret + b[i] * d[x] % mo) % mo;
  }
  return ret;
}
void dfs(int k) {
  l[k] = ++cnt;
  for (int i = 0; i < e[k].size(); ++i) {
    int u = e[k][i];
    d[cnt + 1] = d[l[k]] + 1;
    dfs(u);
  }
  r[k] = cnt;
}
int main() {
  scanf("%d", &n);
  int t, v, x, k;
  for (int i = 2; i <= n; ++i) {
    scanf("%d", &x);
    e[x].push_back(i);
  }
  d[1] = 1;
  dfs(1);
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &v, &x, &k);
      add(l[v], (d[l[v]] * k % mo + x) % mo, -k);
      add(r[v] + 1, -(d[l[v]] * k % mo + x) % mo, k);
    } else {
      scanf("%d", &v);
      int ans = get(l[v]);
      if (ans < 0) ans += mo;
      printf("%d\n", ans);
    }
  }
  return 0;
}

```