#include <bits/stdc++.h>
using namespace std;
const long long maxN = 3e5 + 13, mod = 1e9 + 7;
long long n, q, st[maxN], en[maxN], lvl[maxN], t;
vector<long long> adj[maxN];
void dfs(long long x) {
  st[x] = t++;
  for (auto a : adj[x]) {
    lvl[a] = lvl[x] + 1;
    dfs(a);
  }
  en[x] = t;
}
struct seg_tree {
  long long tree[maxN * 4], lzy[maxN * 4];
  void relax(long long l, long long r, long long indx) {
    if (lzy[indx]) {
      tree[indx] = (tree[indx] + lzy[indx]) % mod;
      if (r - l > 1) {
        lzy[indx << 1] = (lzy[indx << 1] + lzy[indx]) % mod;
        lzy[indx << 1 ^ 1] = (lzy[indx << 1 ^ 1] + lzy[indx]) % mod;
      }
      lzy[indx] = 0;
    }
  }
  void update(long long L, long long R, long long val, long long l = 0,
              long long r = n, long long indx = 1) {
    relax(l, r, indx);
    if (L <= l && r <= R) {
      lzy[indx] = (lzy[indx] + val) % mod;
      relax(l, r, indx);
      return;
    }
    if (R <= l || r <= L) return;
    long long mid = l + r >> 1;
    update(L, R, val, l, mid, indx << 1);
    update(L, R, val, mid, r, indx << 1 ^ 1);
    tree[indx] = (tree[indx << 1] + tree[indx << 1 ^ 1]) % mod;
  }
  long long get(long long L, long long R, long long l = 0, long long r = n,
                long long indx = 1) {
    relax(l, r, indx);
    if (L <= l && r <= R) {
      return tree[indx];
    }
    if (R <= l || r <= L) return 0;
    long long mid = l + r >> 1;
    return (get(L, R, l, mid, indx << 1) + get(L, R, mid, r, indx << 1 ^ 1)) %
           mod;
  }
} seg[5];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    adj[p].push_back(i);
  }
  lvl[1] = 1;
  dfs(1);
  cin >> q;
  while (q--) {
    long long t, v, x, k;
    cin >> t;
    if (t == 1) {
      cin >> v >> x >> k;
      seg[1].update(st[v], en[v], x + lvl[v] * k);
      seg[2].update(st[v], en[v], k);
    } else {
      cin >> v;
      cout << (((seg[1].get(st[v], st[v] + 1) -
                 ((1ll * seg[2].get(st[v], st[v] + 1) * lvl[v]) % mod)) %
                mod) +
               mod) %
                  mod
           << '\n';
    }
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 maxN = 3e5 + 13, mod = 1e9 + 7;
long long n, q, st[maxN], en[maxN], lvl[maxN], t;
vector<long long> adj[maxN];
void dfs(long long x) {
  st[x] = t++;
  for (auto a : adj[x]) {
    lvl[a] = lvl[x] + 1;
    dfs(a);
  }
  en[x] = t;
}
struct seg_tree {
  long long tree[maxN * 4], lzy[maxN * 4];
  void relax(long long l, long long r, long long indx) {
    if (lzy[indx]) {
      tree[indx] = (tree[indx] + lzy[indx]) % mod;
      if (r - l > 1) {
        lzy[indx << 1] = (lzy[indx << 1] + lzy[indx]) % mod;
        lzy[indx << 1 ^ 1] = (lzy[indx << 1 ^ 1] + lzy[indx]) % mod;
      }
      lzy[indx] = 0;
    }
  }
  void update(long long L, long long R, long long val, long long l = 0,
              long long r = n, long long indx = 1) {
    relax(l, r, indx);
    if (L <= l && r <= R) {
      lzy[indx] = (lzy[indx] + val) % mod;
      relax(l, r, indx);
      return;
    }
    if (R <= l || r <= L) return;
    long long mid = l + r >> 1;
    update(L, R, val, l, mid, indx << 1);
    update(L, R, val, mid, r, indx << 1 ^ 1);
    tree[indx] = (tree[indx << 1] + tree[indx << 1 ^ 1]) % mod;
  }
  long long get(long long L, long long R, long long l = 0, long long r = n,
                long long indx = 1) {
    relax(l, r, indx);
    if (L <= l && r <= R) {
      return tree[indx];
    }
    if (R <= l || r <= L) return 0;
    long long mid = l + r >> 1;
    return (get(L, R, l, mid, indx << 1) + get(L, R, mid, r, indx << 1 ^ 1)) %
           mod;
  }
} seg[5];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    adj[p].push_back(i);
  }
  lvl[1] = 1;
  dfs(1);
  cin >> q;
  while (q--) {
    long long t, v, x, k;
    cin >> t;
    if (t == 1) {
      cin >> v >> x >> k;
      seg[1].update(st[v], en[v], x + lvl[v] * k);
      seg[2].update(st[v], en[v], k);
    } else {
      cin >> v;
      cout << (((seg[1].get(st[v], st[v] + 1) -
                 ((1ll * seg[2].get(st[v], st[v] + 1) * lvl[v]) % mod)) %
                mod) +
               mod) %
                  mod
           << '\n';
    }
  }
}

```