#include <bits/stdc++.h>
using namespace std;
const long long M = 1000000007;
vector<int> g[300007];
int parent[300007];
int dc = 0;
int lvl[300007];
int first_sgm[300007], last_sgm[300007];
int ccc = 0;
inline void m_add(long long& a, long long b) {
  a += b;
  if (a >= M)
    a -= M;
  else if (a < 0)
    a += M;
}
template <typename T>
class segment_tree {
 private:
  vector<T> tree;
  int n, sz;

 public:
  segment_tree(int _size) {
    n = _size;
    sz = 1;
    while (sz < _size) sz <<= 1;
    sz <<= 1;
    tree = vector<T>(sz);
  }
  void add(int left, int right, T val) { _add(1, 0, n, left, right + 1, val); }
  T get(int index) { return _get(1, 0, n, index); }

 private:
  void _add(int v, int tl, int tr, int l, int r, T val) {
    if (l >= r) {
      return;
    }
    if (l == tl && r == tr) {
      m_add(tree[v], val);
      return;
    }
    int tm = (tl + tr) / 2;
    _add(v * 2, tl, tm, l, min(r, tm), val);
    _add(v * 2 + 1, tm, tr, max(l, tm), r, val);
  }
  T _get(int v, int tl, int tr, int x) {
    if (tl == x && tr == x + 1) return tree[v];
    int tm = (tl + tr) / 2;
    if (x < tm)
      return tree[v] + _get(2 * v, tl, tm, x);
    else
      return tree[v] + _get(2 * v + 1, tm, tr, x);
  }
};
void dfs(int v, int lev) {
  lvl[v] = lev;
  first_sgm[v] = dc++;
  for (vector<int>::const_iterator it = g[v].begin(); it != g[v].end(); it++) {
    dfs(*it, lev + 1);
  }
  last_sgm[v] = dc - 1;
}
int main() {
  int n, q, t, v;
  long long x, k;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &parent[i]);
    g[parent[i]].push_back(i);
  }
  dfs(1, 0);
  segment_tree<long long> st1(n);
  segment_tree<long long> st2(n);
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d", &t, &v);
    if (t == 1) {
      cin >> x >> k;
      st1.add(first_sgm[v], last_sgm[v], (x + lvl[v] * k) % M);
      st2.add(first_sgm[v], last_sgm[v], k);
    } else if (t == 2) {
      int i = first_sgm[v];
      long long ans = st1.get(i) - st2.get(i) * lvl[v];
      ans = ans % M;
      if (ans < 0) ans += M;
      cout << ans << "\n";
    }
  }
  return 0;
}

### 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 long long M = 1000000007;
vector<int> g[300007];
int parent[300007];
int dc = 0;
int lvl[300007];
int first_sgm[300007], last_sgm[300007];
int ccc = 0;
inline void m_add(long long& a, long long b) {
  a += b;
  if (a >= M)
    a -= M;
  else if (a < 0)
    a += M;
}
template <typename T>
class segment_tree {
 private:
  vector<T> tree;
  int n, sz;

 public:
  segment_tree(int _size) {
    n = _size;
    sz = 1;
    while (sz < _size) sz <<= 1;
    sz <<= 1;
    tree = vector<T>(sz);
  }
  void add(int left, int right, T val) { _add(1, 0, n, left, right + 1, val); }
  T get(int index) { return _get(1, 0, n, index); }

 private:
  void _add(int v, int tl, int tr, int l, int r, T val) {
    if (l >= r) {
      return;
    }
    if (l == tl && r == tr) {
      m_add(tree[v], val);
      return;
    }
    int tm = (tl + tr) / 2;
    _add(v * 2, tl, tm, l, min(r, tm), val);
    _add(v * 2 + 1, tm, tr, max(l, tm), r, val);
  }
  T _get(int v, int tl, int tr, int x) {
    if (tl == x && tr == x + 1) return tree[v];
    int tm = (tl + tr) / 2;
    if (x < tm)
      return tree[v] + _get(2 * v, tl, tm, x);
    else
      return tree[v] + _get(2 * v + 1, tm, tr, x);
  }
};
void dfs(int v, int lev) {
  lvl[v] = lev;
  first_sgm[v] = dc++;
  for (vector<int>::const_iterator it = g[v].begin(); it != g[v].end(); it++) {
    dfs(*it, lev + 1);
  }
  last_sgm[v] = dc - 1;
}
int main() {
  int n, q, t, v;
  long long x, k;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &parent[i]);
    g[parent[i]].push_back(i);
  }
  dfs(1, 0);
  segment_tree<long long> st1(n);
  segment_tree<long long> st2(n);
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d", &t, &v);
    if (t == 1) {
      cin >> x >> k;
      st1.add(first_sgm[v], last_sgm[v], (x + lvl[v] * k) % M);
      st2.add(first_sgm[v], last_sgm[v], k);
    } else if (t == 2) {
      int i = first_sgm[v];
      long long ans = st1.get(i) - st2.get(i) * lvl[v];
      ans = ans % M;
      if (ans < 0) ans += M;
      cout << ans << "\n";
    }
  }
  return 0;
}

```