#include <bits/stdc++.h>
using namespace std;
const int maxn = 601005;
const long long mod = (long long)1e9 + 7;
long long tree[2][maxn];
long long modd(long long x) { return ((x % mod) + mod) % mod; }
void update(int i, int idx, long long val) {
  while (idx < maxn) {
    tree[i][idx] = modd(tree[i][idx] + val);
    idx += idx & (-idx);
  }
}
long long query(int i, int idx) {
  long long sum = 0;
  while (idx) {
    sum = modd(sum + tree[i][idx]);
    idx -= idx & (-idx);
  }
  return sum;
}
int cnt = 0;
int in[maxn], out[maxn], lvl[maxn];
vector<int> graph[maxn];
void dfs(int u, int p) {
  in[u] = ++cnt;
  lvl[u] = lvl[p] + 1;
  for (auto &v : graph[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
  out[u] = cnt;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    graph[p].push_back(i);
    graph[i].push_back(p);
  }
  lvl[0] = -1;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int op, v;
    long long x, k;
    scanf("%d %d", &op, &v);
    if (op == 1) {
      scanf("%lld %lld", &x, &k);
      long long a = modd(x + modd(lvl[v] * k));
      update(0, in[v], a);
      update(0, out[v] + 1, -a);
      update(1, in[v], k);
      update(1, out[v] + 1, -k);
    } else {
      long long ans = query(0, in[v]);
      ans = modd(ans - modd(lvl[v] * query(1, in[v])));
      printf("%lld\n", ans);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 int maxn = 601005;
const long long mod = (long long)1e9 + 7;
long long tree[2][maxn];
long long modd(long long x) { return ((x % mod) + mod) % mod; }
void update(int i, int idx, long long val) {
  while (idx < maxn) {
    tree[i][idx] = modd(tree[i][idx] + val);
    idx += idx & (-idx);
  }
}
long long query(int i, int idx) {
  long long sum = 0;
  while (idx) {
    sum = modd(sum + tree[i][idx]);
    idx -= idx & (-idx);
  }
  return sum;
}
int cnt = 0;
int in[maxn], out[maxn], lvl[maxn];
vector<int> graph[maxn];
void dfs(int u, int p) {
  in[u] = ++cnt;
  lvl[u] = lvl[p] + 1;
  for (auto &v : graph[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
  out[u] = cnt;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    graph[p].push_back(i);
    graph[i].push_back(p);
  }
  lvl[0] = -1;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int op, v;
    long long x, k;
    scanf("%d %d", &op, &v);
    if (op == 1) {
      scanf("%lld %lld", &x, &k);
      long long a = modd(x + modd(lvl[v] * k));
      update(0, in[v], a);
      update(0, out[v] + 1, -a);
      update(1, in[v], k);
      update(1, out[v] + 1, -k);
    } else {
      long long ans = query(0, in[v]);
      ans = modd(ans - modd(lvl[v] * query(1, in[v])));
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```