#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 3e5 + 10;
long long st1[4 * N];
long long st2[4 * N];
int n, m, height[N], s[N], f[N], cur;
vector<int> child[N];
void dfs(int v) {
  s[v] = ++cur;
  for (int u : child[v]) {
    height[u] = height[v] + 1;
    dfs(u);
  }
  f[v] = cur;
}
void update(int v, int x, int k, int pos, int l, int r) {
  if (l > f[v] || r < s[v]) {
    return;
  }
  if (l >= s[v] && r <= f[v]) {
    (st1[pos] += x + 1LL * k * height[v]) %= MOD;
    (st2[pos] += k) %= MOD;
    return;
  }
  int m = (l + r) / 2;
  update(v, x, k, 2 * pos, l, m);
  update(v, x, k, 2 * pos + 1, m + 1, r);
}
long long get(int v, int pos, int l, int r) {
  long long res = (st1[pos] - (height[v] * st2[pos]) % MOD + MOD) % MOD;
  if (l == r) {
    return res;
  }
  int m = (l + r) / 2;
  if (s[v] <= m) {
    return (res + get(v, 2 * pos, l, m)) % MOD;
  } else {
    return (res + get(v, 2 * pos + 1, m + 1, r)) % MOD;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int parent;
    cin >> parent;
    child[parent].push_back(i);
  }
  dfs(1);
  cin >> m;
  for (int i = 0; i < m; i++) {
    int type;
    int v;
    cin >> type >> v;
    if (type == 1) {
      int x, k;
      cin >> x >> k;
      update(v, x, k, 1, 1, n);
    }
    if (type == 2) {
      cout << get(v, 1, 1, n) << endl;
    }
  }
}

### 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 int MOD = 1e9 + 7;
const int N = 3e5 + 10;
long long st1[4 * N];
long long st2[4 * N];
int n, m, height[N], s[N], f[N], cur;
vector<int> child[N];
void dfs(int v) {
  s[v] = ++cur;
  for (int u : child[v]) {
    height[u] = height[v] + 1;
    dfs(u);
  }
  f[v] = cur;
}
void update(int v, int x, int k, int pos, int l, int r) {
  if (l > f[v] || r < s[v]) {
    return;
  }
  if (l >= s[v] && r <= f[v]) {
    (st1[pos] += x + 1LL * k * height[v]) %= MOD;
    (st2[pos] += k) %= MOD;
    return;
  }
  int m = (l + r) / 2;
  update(v, x, k, 2 * pos, l, m);
  update(v, x, k, 2 * pos + 1, m + 1, r);
}
long long get(int v, int pos, int l, int r) {
  long long res = (st1[pos] - (height[v] * st2[pos]) % MOD + MOD) % MOD;
  if (l == r) {
    return res;
  }
  int m = (l + r) / 2;
  if (s[v] <= m) {
    return (res + get(v, 2 * pos, l, m)) % MOD;
  } else {
    return (res + get(v, 2 * pos + 1, m + 1, r)) % MOD;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int parent;
    cin >> parent;
    child[parent].push_back(i);
  }
  dfs(1);
  cin >> m;
  for (int i = 0; i < m; i++) {
    int type;
    int v;
    cin >> type >> v;
    if (type == 1) {
      int x, k;
      cin >> x >> k;
      update(v, x, k, 1, 1, n);
    }
    if (type == 2) {
      cout << get(v, 1, 1, n) << endl;
    }
  }
}

```