#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 3e5 + 100;
vector<int> adj[maxn];
int tin[maxn], tout[maxn], depth[maxn];
long long t1[maxn * 4], t2[maxn * 4];
int n, q, timer = 1;
void dfs(int v, int par = -1) {
  tin[v] = timer++;
  if (par != -1) depth[v] = depth[par] + 1;
  for (auto to : adj[v]) {
    if (to == par) continue;
    if (!tin[to]) dfs(to, v);
  }
  tout[v] = timer;
}
void modify(int v, long long x, long long k, int id = 1, int l = 0, int r = n) {
  if (l >= tout[v] - 1 || r <= tin[v] - 1) return;
  if (l >= tin[v] - 1 && r <= tout[v] - 1) {
    t1[id] = (t1[id] + x) % mod;
    long long add = (depth[v] * k) % mod;
    t1[id] = (t1[id] + add) % mod;
    t2[id] = (t2[id] + k) % mod;
    return;
  }
  int mid = (l + r) / 2;
  modify(v, x, k, id * 2, l, mid);
  modify(v, x, k, id * 2 + 1, mid, r);
}
long long query(int v, int id = 1, int l = 0, int r = n) {
  long long add = (depth[v] * t2[id]) % mod;
  long long tot = (t1[id] + mod - add) % mod;
  if (r - l < 2) return tot;
  int mid = (l + r) / 2;
  if (tin[v] - 1 < mid)
    return (tot + query(v, id * 2, l, mid)) % mod;
  else
    return (tot + query(v, id * 2 + 1, mid, r)) % mod;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int a;
    cin >> a;
    adj[a].push_back(i);
  }
  dfs(1);
  cin >> q;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int v, x, k;
      cin >> v >> x >> k;
      modify(v, x, k);
    } else {
      int v;
      cin >> v;
      cout << query(v) << "\n";
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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 maxn = 3e5 + 100;
vector<int> adj[maxn];
int tin[maxn], tout[maxn], depth[maxn];
long long t1[maxn * 4], t2[maxn * 4];
int n, q, timer = 1;
void dfs(int v, int par = -1) {
  tin[v] = timer++;
  if (par != -1) depth[v] = depth[par] + 1;
  for (auto to : adj[v]) {
    if (to == par) continue;
    if (!tin[to]) dfs(to, v);
  }
  tout[v] = timer;
}
void modify(int v, long long x, long long k, int id = 1, int l = 0, int r = n) {
  if (l >= tout[v] - 1 || r <= tin[v] - 1) return;
  if (l >= tin[v] - 1 && r <= tout[v] - 1) {
    t1[id] = (t1[id] + x) % mod;
    long long add = (depth[v] * k) % mod;
    t1[id] = (t1[id] + add) % mod;
    t2[id] = (t2[id] + k) % mod;
    return;
  }
  int mid = (l + r) / 2;
  modify(v, x, k, id * 2, l, mid);
  modify(v, x, k, id * 2 + 1, mid, r);
}
long long query(int v, int id = 1, int l = 0, int r = n) {
  long long add = (depth[v] * t2[id]) % mod;
  long long tot = (t1[id] + mod - add) % mod;
  if (r - l < 2) return tot;
  int mid = (l + r) / 2;
  if (tin[v] - 1 < mid)
    return (tot + query(v, id * 2, l, mid)) % mod;
  else
    return (tot + query(v, id * 2 + 1, mid, r)) % mod;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int a;
    cin >> a;
    adj[a].push_back(i);
  }
  dfs(1);
  cin >> q;
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int v, x, k;
      cin >> v >> x >> k;
      modify(v, x, k);
    } else {
      int v;
      cin >> v;
      cout << query(v) << "\n";
    }
  }
  return 0;
}

```