#include <bits/stdc++.h>
using namespace std;
const int N = (int)3e5 + 5;
const long long MOD = (long long)1e9 + 7;
long long BIT[2][N];
int in[N], ot[N];
int depth[N];
int cnt;
vector<int> chld[N];
int n, q;
void dfs(int u) {
  in[u] = ++cnt;
  for (int v : chld[u]) {
    depth[v] = depth[u] + 1;
    dfs(v);
  }
  ot[u] = cnt;
}
void update(int id, int x, long long val) {
  val %= MOD;
  for (; x <= n; x += x & -x) BIT[id][x] = (BIT[id][x] + val) % MOD;
}
long long query(int id, int x) {
  long long res = 0;
  while (x) {
    res += BIT[id][x];
    res %= MOD;
    x -= x & -x;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int papa;
    scanf("%d", &papa);
    chld[papa].push_back(i);
  }
  dfs(1);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int opt, v;
    scanf("%d %d", &opt, &v);
    int l = in[v], r = ot[v];
    if (opt == 2) {
      long long res = query(0, l);
      res += (query(1, l)) * (long long)depth[v];
      res %= MOD;
      if (res < 0) res += MOD;
      printf("%lld\n", res);
    } else {
      long long k, x;
      scanf("%lld %lld", &x, &k);
      update(0, l, x + k * (long long)depth[v]);
      update(0, r + 1, -(x + k * (long long)depth[v]));
      update(1, l, -k);
      update(1, r + 1, 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;
const int N = (int)3e5 + 5;
const long long MOD = (long long)1e9 + 7;
long long BIT[2][N];
int in[N], ot[N];
int depth[N];
int cnt;
vector<int> chld[N];
int n, q;
void dfs(int u) {
  in[u] = ++cnt;
  for (int v : chld[u]) {
    depth[v] = depth[u] + 1;
    dfs(v);
  }
  ot[u] = cnt;
}
void update(int id, int x, long long val) {
  val %= MOD;
  for (; x <= n; x += x & -x) BIT[id][x] = (BIT[id][x] + val) % MOD;
}
long long query(int id, int x) {
  long long res = 0;
  while (x) {
    res += BIT[id][x];
    res %= MOD;
    x -= x & -x;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int papa;
    scanf("%d", &papa);
    chld[papa].push_back(i);
  }
  dfs(1);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int opt, v;
    scanf("%d %d", &opt, &v);
    int l = in[v], r = ot[v];
    if (opt == 2) {
      long long res = query(0, l);
      res += (query(1, l)) * (long long)depth[v];
      res %= MOD;
      if (res < 0) res += MOD;
      printf("%lld\n", res);
    } else {
      long long k, x;
      scanf("%lld %lld", &x, &k);
      update(0, l, x + k * (long long)depth[v]);
      update(0, r + 1, -(x + k * (long long)depth[v]));
      update(1, l, -k);
      update(1, r + 1, k);
    }
  }
  return 0;
}

```