#include <bits/stdc++.h>
using namespace std;
const long long wty = 1000000007;
const int N = 3e5 + 5;
int n;
int seg[N], size[N], dep[N], cnt = 0;
class tree {
 private:
  long long num[N];
  int lowbit(int x) { return (x & (-x)); }

 public:
  void update(int u, int val) {
    while (u <= n) {
      num[u] = (num[u] + val) % wty;
      u += lowbit(u);
    }
  }
  void add(int l, int r, long long val) {
    update(l, val);
    update(r + 1, -val);
  }
  long long query(int u) {
    long long res = 0;
    while (u) {
      res = (res + num[u]) % wty;
      u -= lowbit(u);
    }
    return (res + wty) % wty;
  }
} t1, t2;
struct Edge {
  int v, nxt;
} edge[N * 2];
int head[N], k = 1;
void addedge(int u, int v) {
  edge[++k].v = v;
  edge[k].nxt = head[u];
  head[u] = k;
}
void dfs(int u, int fa) {
  seg[u] = ++cnt;
  size[u] = 1;
  dep[u] = dep[fa] + 1;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].v;
    if (v == fa) continue;
    dfs(v, u);
    size[u] += size[v];
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int fa;
    cin >> fa;
    addedge(fa, i), addedge(i, fa);
  }
  dfs(1, 0);
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int opt;
    cin >> opt;
    if (opt == 1) {
      long long u, x, y;
      cin >> u >> x >> y;
      t1.add(seg[u], seg[u] + size[u] - 1, (x + dep[u] * y % wty));
      t2.add(seg[u], seg[u] + size[u] - 1, -y);
    } else {
      int u;
      cin >> u;
      cout << (t1.query(seg[u]) + t2.query(seg[u]) * dep[u] % wty + wty) % wty
           << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate 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 long long wty = 1000000007;
const int N = 3e5 + 5;
int n;
int seg[N], size[N], dep[N], cnt = 0;
class tree {
 private:
  long long num[N];
  int lowbit(int x) { return (x & (-x)); }

 public:
  void update(int u, int val) {
    while (u <= n) {
      num[u] = (num[u] + val) % wty;
      u += lowbit(u);
    }
  }
  void add(int l, int r, long long val) {
    update(l, val);
    update(r + 1, -val);
  }
  long long query(int u) {
    long long res = 0;
    while (u) {
      res = (res + num[u]) % wty;
      u -= lowbit(u);
    }
    return (res + wty) % wty;
  }
} t1, t2;
struct Edge {
  int v, nxt;
} edge[N * 2];
int head[N], k = 1;
void addedge(int u, int v) {
  edge[++k].v = v;
  edge[k].nxt = head[u];
  head[u] = k;
}
void dfs(int u, int fa) {
  seg[u] = ++cnt;
  size[u] = 1;
  dep[u] = dep[fa] + 1;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].v;
    if (v == fa) continue;
    dfs(v, u);
    size[u] += size[v];
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int fa;
    cin >> fa;
    addedge(fa, i), addedge(i, fa);
  }
  dfs(1, 0);
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int opt;
    cin >> opt;
    if (opt == 1) {
      long long u, x, y;
      cin >> u >> x >> y;
      t1.add(seg[u], seg[u] + size[u] - 1, (x + dep[u] * y % wty));
      t2.add(seg[u], seg[u] + size[u] - 1, -y);
    } else {
      int u;
      cin >> u;
      cout << (t1.query(seg[u]) + t2.query(seg[u]) * dep[u] % wty + wty) % wty
           << endl;
    }
  }
  return 0;
}

```