#include <bits/stdc++.h>
using namespace std;
template <typename T = int, T _mod = 1000000007, typename U = long long>
struct Minter {
  static constexpr int mod = _mod;
  T val;
  Minter() : val(0) {}
  Minter(U x) {
    x %= mod;
    if (x < 0) x += mod;
    val = x;
  }
  explicit operator int() const { return val; }
  Minter& operator+=(Minter oth) {
    val += oth.val;
    if (val >= mod) val -= mod;
    return *this;
  }
  Minter& operator-=(Minter oth) {
    val -= oth.val;
    if (val < 0) val += mod;
    return *this;
  }
  Minter& operator*=(Minter oth) {
    val = ((U)val) * oth.val % mod;
    return *this;
  }
  Minter& operator/=(Minter oth) { return *this *= oth.inv(); }
  Minter& operator^=(U oth) { return *this = pwr(*this, oth); }
  Minter operator+(const Minter& oth) const { return Minter(*this) += oth; }
  Minter operator-(const Minter& oth) const { return Minter(*this) -= oth; }
  Minter operator*(const Minter& oth) const { return Minter(*this) *= oth; }
  Minter operator/(const Minter& oth) const { return Minter(*this) /= oth; }
  Minter operator^(long long oth) const { return Minter(*this) ^= oth; }
  bool operator<(Minter oth) const { return val < oth.val; }
  bool operator>(Minter oth) const { return val > oth.val; }
  bool operator<=(Minter oth) const { return val <= oth.val; }
  bool operator>=(Minter oth) const { return val >= oth.val; }
  bool operator==(Minter oth) const { return val == oth.val; }
  bool operator!=(Minter oth) const { return val != oth.val; }
  Minter pwr(Minter a, long long b) const {
    auto r = (Minter)1;
    for (; b; a *= a, b >>= 1)
      if (b & 1) r *= a;
    return r;
  }
};
using U = Minter<>;
ostream& operator<<(ostream& out, const U& oth) { return out << oth.val; }
istream& operator>>(istream& in, U& oth) {
  in >> oth.val;
  return in;
}
const int N = 6e5 + 2;
int p[N], d[N];
U s1[4 * N], s2[4 * N], lazy1[4 * N], lazy2[4 * N];
int n, q, timer = -1;
vector<int> adj[N];
int Tstart[N], Tend[N], vis[N];
void dfs(int v, int dep) {
  vis[v] = 1;
  d[v] = dep + 1;
  Tstart[v] = ++timer;
  for (auto x : adj[v]) {
    if (!vis[x]) {
      dfs(x, dep + 1);
    }
  }
  Tend[v] = ++timer;
}
void init() {
  for (int i = 0; i < n - 1; i++) {
    cin >> p[i];
    adj[p[i]].push_back(i + 2);
    adj[i + 2].push_back(p[i]);
  }
  dfs(1, 0);
}
void change(int id, int l, int r, U x1, U x2) {
  lazy1[id] += x1;
  s1[id] += x1;
  lazy2[id] += x2;
  s2[id] += x2;
}
void pass(int id, int l, int r) {
  int mid = (l + r) / 2;
  change(2 * id, l, mid, lazy1[id], lazy2[id]);
  change(2 * id + 1, l, mid, lazy1[id], lazy2[id]);
  lazy1[id] = lazy2[id] = 0;
}
void update(int L, int R, U x1, U x2, int id = 1, int l = 0, int r = timer) {
  if (R < l or L > r) {
    return;
  }
  if (L <= l and r <= R) {
    change(id, l, r, x1, x2);
    return;
  }
  pass(id, l, r);
  int mid = (l + r) / 2;
  update(L, R, x1, x2, 2 * id, l, mid);
  update(L, R, x1, x2, 2 * id + 1, mid + 1, r);
  s1[id] = s1[2 * id] + s1[2 * id + 1];
  s2[id] = s2[2 * id] + s2[2 * id + 1];
}
U query(int L, int R, int v, int id = 1, int l = 0, int r = timer) {
  if (R < l or L > r) {
    return (U)0;
  }
  if (L <= l and r <= R) {
    return (U)(s1[id] - s2[id] * d[v]);
  }
  pass(id, l, r);
  int mid = (l + r) / 2;
  return (query(L, R, v, 2 * id, l, mid) +
          query(L, R, v, 2 * id + 1, mid + 1, r));
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  init();
  cin >> q;
  while (q--) {
    int t, v;
    cin >> t >> v;
    if (t == 2) {
      cout << query(Tstart[v], Tstart[v], v) << "\n";
      continue;
    }
    U x, k;
    cin >> x >> k;
    update(Tstart[v], Tend[v] - 1, x + (k * d[v]), k);
  }
  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;
template <typename T = int, T _mod = 1000000007, typename U = long long>
struct Minter {
  static constexpr int mod = _mod;
  T val;
  Minter() : val(0) {}
  Minter(U x) {
    x %= mod;
    if (x < 0) x += mod;
    val = x;
  }
  explicit operator int() const { return val; }
  Minter& operator+=(Minter oth) {
    val += oth.val;
    if (val >= mod) val -= mod;
    return *this;
  }
  Minter& operator-=(Minter oth) {
    val -= oth.val;
    if (val < 0) val += mod;
    return *this;
  }
  Minter& operator*=(Minter oth) {
    val = ((U)val) * oth.val % mod;
    return *this;
  }
  Minter& operator/=(Minter oth) { return *this *= oth.inv(); }
  Minter& operator^=(U oth) { return *this = pwr(*this, oth); }
  Minter operator+(const Minter& oth) const { return Minter(*this) += oth; }
  Minter operator-(const Minter& oth) const { return Minter(*this) -= oth; }
  Minter operator*(const Minter& oth) const { return Minter(*this) *= oth; }
  Minter operator/(const Minter& oth) const { return Minter(*this) /= oth; }
  Minter operator^(long long oth) const { return Minter(*this) ^= oth; }
  bool operator<(Minter oth) const { return val < oth.val; }
  bool operator>(Minter oth) const { return val > oth.val; }
  bool operator<=(Minter oth) const { return val <= oth.val; }
  bool operator>=(Minter oth) const { return val >= oth.val; }
  bool operator==(Minter oth) const { return val == oth.val; }
  bool operator!=(Minter oth) const { return val != oth.val; }
  Minter pwr(Minter a, long long b) const {
    auto r = (Minter)1;
    for (; b; a *= a, b >>= 1)
      if (b & 1) r *= a;
    return r;
  }
};
using U = Minter<>;
ostream& operator<<(ostream& out, const U& oth) { return out << oth.val; }
istream& operator>>(istream& in, U& oth) {
  in >> oth.val;
  return in;
}
const int N = 6e5 + 2;
int p[N], d[N];
U s1[4 * N], s2[4 * N], lazy1[4 * N], lazy2[4 * N];
int n, q, timer = -1;
vector<int> adj[N];
int Tstart[N], Tend[N], vis[N];
void dfs(int v, int dep) {
  vis[v] = 1;
  d[v] = dep + 1;
  Tstart[v] = ++timer;
  for (auto x : adj[v]) {
    if (!vis[x]) {
      dfs(x, dep + 1);
    }
  }
  Tend[v] = ++timer;
}
void init() {
  for (int i = 0; i < n - 1; i++) {
    cin >> p[i];
    adj[p[i]].push_back(i + 2);
    adj[i + 2].push_back(p[i]);
  }
  dfs(1, 0);
}
void change(int id, int l, int r, U x1, U x2) {
  lazy1[id] += x1;
  s1[id] += x1;
  lazy2[id] += x2;
  s2[id] += x2;
}
void pass(int id, int l, int r) {
  int mid = (l + r) / 2;
  change(2 * id, l, mid, lazy1[id], lazy2[id]);
  change(2 * id + 1, l, mid, lazy1[id], lazy2[id]);
  lazy1[id] = lazy2[id] = 0;
}
void update(int L, int R, U x1, U x2, int id = 1, int l = 0, int r = timer) {
  if (R < l or L > r) {
    return;
  }
  if (L <= l and r <= R) {
    change(id, l, r, x1, x2);
    return;
  }
  pass(id, l, r);
  int mid = (l + r) / 2;
  update(L, R, x1, x2, 2 * id, l, mid);
  update(L, R, x1, x2, 2 * id + 1, mid + 1, r);
  s1[id] = s1[2 * id] + s1[2 * id + 1];
  s2[id] = s2[2 * id] + s2[2 * id + 1];
}
U query(int L, int R, int v, int id = 1, int l = 0, int r = timer) {
  if (R < l or L > r) {
    return (U)0;
  }
  if (L <= l and r <= R) {
    return (U)(s1[id] - s2[id] * d[v]);
  }
  pass(id, l, r);
  int mid = (l + r) / 2;
  return (query(L, R, v, 2 * id, l, mid) +
          query(L, R, v, 2 * id + 1, mid + 1, r));
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  init();
  cin >> q;
  while (q--) {
    int t, v;
    cin >> t >> v;
    if (t == 2) {
      cout << query(Tstart[v], Tstart[v], v) << "\n";
      continue;
    }
    U x, k;
    cin >> x >> k;
    update(Tstart[v], Tend[v] - 1, x + (k * d[v]), k);
  }
  return 0;
}

```