#include <bits/stdc++.h>
using namespace std;
const double pi = 3.1415926535897932384626433832795;
const int inf = 1e9;
const long long inf2 = 1e18;
const int mod = 1e9 + 7;
void boost() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
const int N = 3e5 + 2;
vector<int> adj[N];
long long tree[2][2 * N];
int tin[N], tout[N], timer = 1;
long long depth[N];
void dfs(int u, int d) {
  tin[u] = timer++;
  depth[u] = d;
  for (auto v : adj[u]) {
    dfs(v, d + 1);
  }
  tout[u] = timer++;
}
long long read(int idx, int i) {
  long long sum = 0;
  while (idx > 0) {
    sum += tree[i][idx];
    sum %= mod;
    idx -= (idx & -idx);
  }
  return sum;
}
void update(int idx, long long val, int i) {
  while (idx < 2 * N) {
    tree[i][idx] += val;
    tree[i][idx] %= mod;
    idx += (idx & -idx);
  }
}
void solve() {
  int n;
  cin >> n;
  int p;
  for (int i = 2; i < n + 1; ++i) {
    cin >> p;
    adj[p].push_back(i);
  }
  dfs(1, 0);
  int q;
  cin >> q;
  int t, v;
  long long x, k;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> v >> x >> k;
      update(tin[v], x + depth[v] * k, 0);
      update(tout[v], -1LL * (x + depth[v] * k), 0);
      update(tin[v], k, 1);
      update(tout[v], -k, 1);
    } else {
      cin >> v;
      x = read(tin[v], 0);
      x = (x + 100LL * mod - (depth[v] * (read(tin[v], 1))) % mod) % mod;
      cout << x << '\n';
    }
  }
  return;
}
int main() {
  boost();
  int tc = 1;
  while (tc--) solve();
  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 double pi = 3.1415926535897932384626433832795;
const int inf = 1e9;
const long long inf2 = 1e18;
const int mod = 1e9 + 7;
void boost() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
const int N = 3e5 + 2;
vector<int> adj[N];
long long tree[2][2 * N];
int tin[N], tout[N], timer = 1;
long long depth[N];
void dfs(int u, int d) {
  tin[u] = timer++;
  depth[u] = d;
  for (auto v : adj[u]) {
    dfs(v, d + 1);
  }
  tout[u] = timer++;
}
long long read(int idx, int i) {
  long long sum = 0;
  while (idx > 0) {
    sum += tree[i][idx];
    sum %= mod;
    idx -= (idx & -idx);
  }
  return sum;
}
void update(int idx, long long val, int i) {
  while (idx < 2 * N) {
    tree[i][idx] += val;
    tree[i][idx] %= mod;
    idx += (idx & -idx);
  }
}
void solve() {
  int n;
  cin >> n;
  int p;
  for (int i = 2; i < n + 1; ++i) {
    cin >> p;
    adj[p].push_back(i);
  }
  dfs(1, 0);
  int q;
  cin >> q;
  int t, v;
  long long x, k;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> v >> x >> k;
      update(tin[v], x + depth[v] * k, 0);
      update(tout[v], -1LL * (x + depth[v] * k), 0);
      update(tin[v], k, 1);
      update(tout[v], -k, 1);
    } else {
      cin >> v;
      x = read(tin[v], 0);
      x = (x + 100LL * mod - (depth[v] * (read(tin[v], 1))) % mod) % mod;
      cout << x << '\n';
    }
  }
  return;
}
int main() {
  boost();
  int tc = 1;
  while (tc--) solve();
  return 0;
}

```