#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > adj(300005);
long long bit[1000005], bit1[1000005], arr[300005], st[300005], en[300005],
    sz[300005], lev[300005];
long long tim = 0, n;
int dfs(int u, int p, int l) {
  int i;
  lev[u] = l;
  sz[u] = 1;
  arr[tim] = u;
  st[u] = tim;
  tim++;
  for (i = 0; i < adj[u].size(); i++) {
    if (adj[u][i] != p) {
      dfs(adj[u][i], u, l + 1);
      sz[u] += sz[adj[u][i]];
    }
  }
  en[u] = tim - 1;
  return 0;
}
long long update(int pos, long long val) {
  while (pos < n + 10) {
    bit[pos] += val;
    bit[pos] %= (1000 * 1000 * 1000 + 7);
    pos += pos & (-pos);
  }
}
long long update1(int pos, long long val) {
  while (pos < n + 10) {
    bit1[pos] += val;
    bit1[pos] %= (1000 * 1000 * 1000 + 7);
    pos += pos & (-pos);
  }
}
long long query(int pos) {
  long long res = 0;
  while (pos > 0) {
    res += bit[pos];
    res %= (1000 * 1000 * 1000 + 7);
    pos -= pos & (-pos);
  }
  return res;
}
long long query1(int pos) {
  long long res = 0;
  while (pos > 0) {
    res += bit1[pos];
    res %= (1000 * 1000 * 1000 + 7);
    pos -= pos & (-pos);
  }
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  long long i, u, v, typ, x, k, q, val;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    cin >> u;
    u--;
    adj[u].push_back(i + 1);
    adj[i + 1].push_back(u);
  }
  dfs(0, -1, 0);
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> typ >> v;
    v--;
    if (typ == 1) {
      cin >> x >> k;
      update(st[v] + 1, (x + k * lev[v]) % (1000 * 1000 * 1000 + 7));
      update(en[v] + 2, -1 * (x + k * lev[v]) % (1000 * 1000 * 1000 + 7));
      update1(st[v] + 1, k);
      update1(en[v] + 2, -1 * k);
    } else {
      val = query(st[v] + 1);
      val -= query1(st[v] + 1) * lev[v];
      val %= (1000 * 1000 * 1000 + 7);
      val += (1000 * 1000 * 1000 + 7);
      val %= (1000 * 1000 * 1000 + 7);
      cout << val << endl;
    }
  }
}

### 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;
vector<vector<long long> > adj(300005);
long long bit[1000005], bit1[1000005], arr[300005], st[300005], en[300005],
    sz[300005], lev[300005];
long long tim = 0, n;
int dfs(int u, int p, int l) {
  int i;
  lev[u] = l;
  sz[u] = 1;
  arr[tim] = u;
  st[u] = tim;
  tim++;
  for (i = 0; i < adj[u].size(); i++) {
    if (adj[u][i] != p) {
      dfs(adj[u][i], u, l + 1);
      sz[u] += sz[adj[u][i]];
    }
  }
  en[u] = tim - 1;
  return 0;
}
long long update(int pos, long long val) {
  while (pos < n + 10) {
    bit[pos] += val;
    bit[pos] %= (1000 * 1000 * 1000 + 7);
    pos += pos & (-pos);
  }
}
long long update1(int pos, long long val) {
  while (pos < n + 10) {
    bit1[pos] += val;
    bit1[pos] %= (1000 * 1000 * 1000 + 7);
    pos += pos & (-pos);
  }
}
long long query(int pos) {
  long long res = 0;
  while (pos > 0) {
    res += bit[pos];
    res %= (1000 * 1000 * 1000 + 7);
    pos -= pos & (-pos);
  }
  return res;
}
long long query1(int pos) {
  long long res = 0;
  while (pos > 0) {
    res += bit1[pos];
    res %= (1000 * 1000 * 1000 + 7);
    pos -= pos & (-pos);
  }
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  long long i, u, v, typ, x, k, q, val;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    cin >> u;
    u--;
    adj[u].push_back(i + 1);
    adj[i + 1].push_back(u);
  }
  dfs(0, -1, 0);
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> typ >> v;
    v--;
    if (typ == 1) {
      cin >> x >> k;
      update(st[v] + 1, (x + k * lev[v]) % (1000 * 1000 * 1000 + 7));
      update(en[v] + 2, -1 * (x + k * lev[v]) % (1000 * 1000 * 1000 + 7));
      update1(st[v] + 1, k);
      update1(en[v] + 2, -1 * k);
    } else {
      val = query(st[v] + 1);
      val -= query1(st[v] + 1) * lev[v];
      val %= (1000 * 1000 * 1000 + 7);
      val += (1000 * 1000 * 1000 + 7);
      val %= (1000 * 1000 * 1000 + 7);
      cout << val << endl;
    }
  }
}

```