#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
const int mod = 1e9 + 7;
void add(int &x, int v) {
  x += v;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
struct bit {
  int BIT[maxn];
  void update(int x, int v) {
    for (; x < maxn; x += x & -x) add(BIT[x], v);
  }
  int query(int x) {
    int sum = 0;
    for (; x > 0; x -= x & -x) add(sum, BIT[x]);
    return sum;
  }
  void update(int l, int r, int v) {
    update(l, v);
    update(r + 1, -v);
  }
} freq, sum;
int in[maxn], out[maxn], d[maxn], idx = 1, n;
vector<int> adj[maxn];
void dfs(int u, int par) {
  in[u] = idx++;
  for (int v : adj[u])
    if (v - par) {
      d[v] = d[u] + 1;
      dfs(v, u);
    }
  out[u] = idx - 1;
}
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    adj[p].push_back(i);
  }
  dfs(1, 0);
  int q;
  scanf("%d", &q);
  while (q--) {
    int typ, v;
    scanf("%d %d", &typ, &v);
    if (typ == 1) {
      int x, k;
      scanf("%d %d", &x, &k);
      freq.update(in[v], out[v], k);
      sum.update(in[v], out[v], ((d[v] * 1ll * k % mod) + x) % mod);
    } else {
      int ans = sum.query(in[v]);
      ans -= (d[v] * 1ll * freq.query(in[v])) % mod;
      if (ans >= mod) ans -= mod;
      if (ans < 0) ans += mod;
      printf("%d\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;
const int maxn = 3e5 + 10;
const int mod = 1e9 + 7;
void add(int &x, int v) {
  x += v;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
}
struct bit {
  int BIT[maxn];
  void update(int x, int v) {
    for (; x < maxn; x += x & -x) add(BIT[x], v);
  }
  int query(int x) {
    int sum = 0;
    for (; x > 0; x -= x & -x) add(sum, BIT[x]);
    return sum;
  }
  void update(int l, int r, int v) {
    update(l, v);
    update(r + 1, -v);
  }
} freq, sum;
int in[maxn], out[maxn], d[maxn], idx = 1, n;
vector<int> adj[maxn];
void dfs(int u, int par) {
  in[u] = idx++;
  for (int v : adj[u])
    if (v - par) {
      d[v] = d[u] + 1;
      dfs(v, u);
    }
  out[u] = idx - 1;
}
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    adj[p].push_back(i);
  }
  dfs(1, 0);
  int q;
  scanf("%d", &q);
  while (q--) {
    int typ, v;
    scanf("%d %d", &typ, &v);
    if (typ == 1) {
      int x, k;
      scanf("%d %d", &x, &k);
      freq.update(in[v], out[v], k);
      sum.update(in[v], out[v], ((d[v] * 1ll * k % mod) + x) % mod);
    } else {
      int ans = sum.query(in[v]);
      ans -= (d[v] * 1ll * freq.query(in[v])) % mod;
      if (ans >= mod) ans -= mod;
      if (ans < 0) ans += mod;
      printf("%d\n", ans);
    }
  }
}

```