#include <bits/stdc++.h>
using namespace std;
const int maxN = 3e5, MOD = 1e9 + 7;
int st[maxN], en[maxN], lev[maxN], aC = 0, n;
pair<long long, long long> T[maxN << 1];
vector<int> adj[maxN];
void dfs(int node, int le) {
  st[node] = aC++, lev[node] = le;
  for (int i = 0; i < adj[node].size(); i++) dfs(adj[node][i], le + 1);
  en[node] = aC;
}
void apply(int i, long long x, long long k) {
  T[i].first = (T[i].first + x) % MOD, T[i].second = (T[i].second + k) % MOD;
}
void update(int node, long long x, long long k) {
  x = (x + lev[node] * k) % MOD;
  int l = st[node] + n, r = en[node] + n;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) apply(l++, x, k);
    if (r & 1) apply(--r, x, k);
  }
}
long long query(int node) {
  long long x = 0, k = 0, w = st[node] + n;
  while (w) {
    x = (x + T[w].first) % MOD, k = (k + T[w].second) % MOD;
    w >>= 1;
  }
  return (x - (k * lev[node] % MOD) + MOD) % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int v;
    cin >> v;
    adj[--v].push_back(i);
  }
  dfs(0, 0);
  int q;
  cin >> q;
  while (q--) {
    int ty, a, b, c;
    cin >> ty;
    if (ty == 1) {
      cin >> a >> b >> c;
      update(--a, b, c);
    } else {
      cin >> a;
      cout << query(--a) << '\n';
    }
  }
}

### Prompt

Please create a solution in cpp 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 = 3e5, MOD = 1e9 + 7;
int st[maxN], en[maxN], lev[maxN], aC = 0, n;
pair<long long, long long> T[maxN << 1];
vector<int> adj[maxN];
void dfs(int node, int le) {
  st[node] = aC++, lev[node] = le;
  for (int i = 0; i < adj[node].size(); i++) dfs(adj[node][i], le + 1);
  en[node] = aC;
}
void apply(int i, long long x, long long k) {
  T[i].first = (T[i].first + x) % MOD, T[i].second = (T[i].second + k) % MOD;
}
void update(int node, long long x, long long k) {
  x = (x + lev[node] * k) % MOD;
  int l = st[node] + n, r = en[node] + n;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) apply(l++, x, k);
    if (r & 1) apply(--r, x, k);
  }
}
long long query(int node) {
  long long x = 0, k = 0, w = st[node] + n;
  while (w) {
    x = (x + T[w].first) % MOD, k = (k + T[w].second) % MOD;
    w >>= 1;
  }
  return (x - (k * lev[node] % MOD) + MOD) % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int v;
    cin >> v;
    adj[--v].push_back(i);
  }
  dfs(0, 0);
  int q;
  cin >> q;
  while (q--) {
    int ty, a, b, c;
    cin >> ty;
    if (ty == 1) {
      cin >> a >> b >> c;
      update(--a, b, c);
    } else {
      cin >> a;
      cout << query(--a) << '\n';
    }
  }
}

```