#include <bits/stdc++.h>
using namespace std;
vector<int> adj[300006];
int n, depth[300006], in[300006], out[300006], counter = 1;
long long bit[300006][3];
void dfs(int node) {
  in[node] = counter++;
  for (int i = 0; i < adj[node].size(); i++) {
    int to = adj[node][i];
    depth[to] = depth[node] + 1;
    dfs(to);
  }
  out[node] = counter;
}
long long query(int idx, int idx2) {
  long long ret = 0;
  while (idx > 0) {
    ret += bit[idx][idx2];
    ret %= 1000000007;
    idx -= idx & -idx;
  }
  return ret;
}
void update(int idx, long long val, int idx2) {
  while (idx <= n + 1) {
    bit[idx][idx2] += val;
    bit[idx][idx2] %= 1000000007;
    idx += idx & -idx;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int par;
    scanf("%d", &par);
    adj[par].push_back(i);
  }
  depth[1] = 1;
  dfs(1);
  int q;
  scanf("%d", &q);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int v;
      long long x, k;
      scanf("%d%I64d%I64d", &v, &x, &k);
      update(in[v], x + k * depth[v], 0);
      update(out[v], -x - k * depth[v], 0);
      update(in[v], k, 1);
      update(out[v], -k, 1);
    } else {
      int v;
      scanf("%d", &v);
      long long sol =
          (query(in[v], 0) - ((query(in[v], 1) * depth[v]) % 1000000007) +
           1000000007) %
          1000000007;
      while (sol < 0) sol += 1000000007;
      printf("%I64d\n", sol);
    }
  }
  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;
vector<int> adj[300006];
int n, depth[300006], in[300006], out[300006], counter = 1;
long long bit[300006][3];
void dfs(int node) {
  in[node] = counter++;
  for (int i = 0; i < adj[node].size(); i++) {
    int to = adj[node][i];
    depth[to] = depth[node] + 1;
    dfs(to);
  }
  out[node] = counter;
}
long long query(int idx, int idx2) {
  long long ret = 0;
  while (idx > 0) {
    ret += bit[idx][idx2];
    ret %= 1000000007;
    idx -= idx & -idx;
  }
  return ret;
}
void update(int idx, long long val, int idx2) {
  while (idx <= n + 1) {
    bit[idx][idx2] += val;
    bit[idx][idx2] %= 1000000007;
    idx += idx & -idx;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int par;
    scanf("%d", &par);
    adj[par].push_back(i);
  }
  depth[1] = 1;
  dfs(1);
  int q;
  scanf("%d", &q);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int v;
      long long x, k;
      scanf("%d%I64d%I64d", &v, &x, &k);
      update(in[v], x + k * depth[v], 0);
      update(out[v], -x - k * depth[v], 0);
      update(in[v], k, 1);
      update(out[v], -k, 1);
    } else {
      int v;
      scanf("%d", &v);
      long long sol =
          (query(in[v], 0) - ((query(in[v], 1) * depth[v]) % 1000000007) +
           1000000007) %
          1000000007;
      while (sol < 0) sol += 1000000007;
      printf("%I64d\n", sol);
    }
  }
  return 0;
}

```