#include <bits/stdc++.h>
using namespace std;
const int maxN = 3e5 + 10;
const int MOD = 1e9 + 7;
long long data[2][4 * maxN];
long long n;
vector<long long> adj[maxN];
long long st[maxN], fn[maxN], h[maxN];
long long boz;
long long sum(long long a, long long b) { return ((a + b) % MOD + MOD) % MOD; }
void dfs(long long v, long long p = -1) {
  st[v] = boz++;
  if (p != -1) h[v] = h[p] + 1;
  for (long long i = 0; i < adj[v].size(); i++) {
    long long u = adj[v][i];
    if (u != p) dfs(u, v);
  }
  fn[v] = boz;
}
void add(long long h, long long l, long long r, long long x, long long s = 0,
         long long e = n, long long id = 1) {
  if (e <= l || r <= s) return;
  if (l <= s && e <= r) {
    data[h][id] = sum(data[h][id], x);
    return;
  }
  long long mid = s + e >> 1;
  add(h, l, r, x, s, mid, 2 * id);
  add(h, l, r, x, mid, e, 2 * id + 1);
}
long long get(long long h, long long p, long long s = 0, long long e = n,
              long long id = 1) {
  if (e - s < 2) return data[h][id];
  long long mid = s + e >> 1;
  if (p < mid)
    return sum(get(h, p, s, mid, 2 * id), data[h][id]);
  else
    return sum(get(h, p, mid, e, 2 * id + 1), data[h][id]);
}
long long ind[maxN];
int32_t main() {
  scanf("%lld", &n);
  for (long long i = 1; i < n; i++) {
    long long v;
    scanf("%lld", &v);
    v--;
    adj[i].push_back(v);
    adj[v].push_back(i);
  }
  h[0] = 0;
  boz = 0;
  dfs(0);
  vector<pair<long long, long long> > ver;
  for (long long i = 0; i < n; i++) ver.push_back(make_pair(st[i], i));
  sort(ver.begin(), ver.end());
  for (long long i = 0; i < ver.size(); i++) {
    long long v = ver[i].second;
    ind[v] = i;
  }
  long long q;
  scanf("%lld", &q);
  while (q--) {
    long long type;
    scanf("%lld", &type);
    if (type == 1) {
      long long v, x, k;
      scanf("%lld %lld %lld", &v, &x, &k);
      v--;
      add(0, st[v], fn[v], sum(h[v] * k % MOD, x));
      add(1, st[v], fn[v], k);
    }
    if (type == 2) {
      long long v;
      scanf("%lld", &v);
      v--;
      printf("%lld\n", sum(get(0, ind[v]), (-1) * get(1, ind[v]) * h[v]));
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 int maxN = 3e5 + 10;
const int MOD = 1e9 + 7;
long long data[2][4 * maxN];
long long n;
vector<long long> adj[maxN];
long long st[maxN], fn[maxN], h[maxN];
long long boz;
long long sum(long long a, long long b) { return ((a + b) % MOD + MOD) % MOD; }
void dfs(long long v, long long p = -1) {
  st[v] = boz++;
  if (p != -1) h[v] = h[p] + 1;
  for (long long i = 0; i < adj[v].size(); i++) {
    long long u = adj[v][i];
    if (u != p) dfs(u, v);
  }
  fn[v] = boz;
}
void add(long long h, long long l, long long r, long long x, long long s = 0,
         long long e = n, long long id = 1) {
  if (e <= l || r <= s) return;
  if (l <= s && e <= r) {
    data[h][id] = sum(data[h][id], x);
    return;
  }
  long long mid = s + e >> 1;
  add(h, l, r, x, s, mid, 2 * id);
  add(h, l, r, x, mid, e, 2 * id + 1);
}
long long get(long long h, long long p, long long s = 0, long long e = n,
              long long id = 1) {
  if (e - s < 2) return data[h][id];
  long long mid = s + e >> 1;
  if (p < mid)
    return sum(get(h, p, s, mid, 2 * id), data[h][id]);
  else
    return sum(get(h, p, mid, e, 2 * id + 1), data[h][id]);
}
long long ind[maxN];
int32_t main() {
  scanf("%lld", &n);
  for (long long i = 1; i < n; i++) {
    long long v;
    scanf("%lld", &v);
    v--;
    adj[i].push_back(v);
    adj[v].push_back(i);
  }
  h[0] = 0;
  boz = 0;
  dfs(0);
  vector<pair<long long, long long> > ver;
  for (long long i = 0; i < n; i++) ver.push_back(make_pair(st[i], i));
  sort(ver.begin(), ver.end());
  for (long long i = 0; i < ver.size(); i++) {
    long long v = ver[i].second;
    ind[v] = i;
  }
  long long q;
  scanf("%lld", &q);
  while (q--) {
    long long type;
    scanf("%lld", &type);
    if (type == 1) {
      long long v, x, k;
      scanf("%lld %lld %lld", &v, &x, &k);
      v--;
      add(0, st[v], fn[v], sum(h[v] * k % MOD, x));
      add(1, st[v], fn[v], k);
    }
    if (type == 2) {
      long long v;
      scanf("%lld", &v);
      v--;
      printf("%lld\n", sum(get(0, ind[v]), (-1) * get(1, ind[v]) * h[v]));
    }
  }
  return 0;
}

```