#include <bits/stdc++.h>
using namespace std;
mt19937 Rand(time(0));
const int maxn = 3e5 + 5;
const long long maxc = 1e9 + 7;
int h[maxn], n, q, in[maxn], out[maxn], cnt;
vector<int> adj[maxn];
void dfs(int u, int par) {
  in[u] = ++cnt;
  for (auto v : adj[u])
    if (v != par) {
      h[v] = h[u] + 1;
      dfs(v, u);
    }
  out[u] = cnt;
}
struct SegmentTree {
  long long t[maxn << 2];
  void up(int node, int l, int r, int u, int v, long long cur) {
    if (l > v || r < u) return;
    if (l >= u && r <= v) {
      t[node] = (t[node] + cur) % maxc;
      return;
    }
    int mid = (l + r) >> 1;
    up(node << 1, l, mid, u, v, cur);
    up(node << 1 | 1, mid + 1, r, u, v, cur);
  }
  long long get(int node, int l, int r, int u) {
    if (l > u || r < u) return 0;
    if (l == r) return t[node];
    int mid = (l + r) >> 1;
    if (u <= mid)
      return (t[node] + get(node << 1, l, mid, u)) % maxc;
    else
      return (t[node] + get(node << 1 | 1, mid + 1, r, u)) % maxc;
  }
} IT1, IT2;
void query() {
  int type, v, x, k;
  cin >> type >> v;
  if (type == 1) {
    cin >> x >> k;
    long long cur = 1ll * x + 1ll * h[v] * k;
    cur %= maxc;
    IT1.up(1, 1, n, in[v], out[v], cur);
    cur = (-k + maxc) % maxc;
    IT2.up(1, 1, n, in[v], out[v], cur);
  } else {
    long long cur1 = IT1.get(1, 1, n, in[v]);
    long long cur2 = IT2.get(1, 1, n, in[v]);
    cout << (cur1 + cur2 * 1ll * h[v]) % maxc << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int u;
    cin >> u;
    adj[u].push_back(i);
  }
  dfs(1, -1);
  cin >> q;
  while (q--) query();
  return 0;
}

### Prompt

Your task is to create a cpp solution 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;
mt19937 Rand(time(0));
const int maxn = 3e5 + 5;
const long long maxc = 1e9 + 7;
int h[maxn], n, q, in[maxn], out[maxn], cnt;
vector<int> adj[maxn];
void dfs(int u, int par) {
  in[u] = ++cnt;
  for (auto v : adj[u])
    if (v != par) {
      h[v] = h[u] + 1;
      dfs(v, u);
    }
  out[u] = cnt;
}
struct SegmentTree {
  long long t[maxn << 2];
  void up(int node, int l, int r, int u, int v, long long cur) {
    if (l > v || r < u) return;
    if (l >= u && r <= v) {
      t[node] = (t[node] + cur) % maxc;
      return;
    }
    int mid = (l + r) >> 1;
    up(node << 1, l, mid, u, v, cur);
    up(node << 1 | 1, mid + 1, r, u, v, cur);
  }
  long long get(int node, int l, int r, int u) {
    if (l > u || r < u) return 0;
    if (l == r) return t[node];
    int mid = (l + r) >> 1;
    if (u <= mid)
      return (t[node] + get(node << 1, l, mid, u)) % maxc;
    else
      return (t[node] + get(node << 1 | 1, mid + 1, r, u)) % maxc;
  }
} IT1, IT2;
void query() {
  int type, v, x, k;
  cin >> type >> v;
  if (type == 1) {
    cin >> x >> k;
    long long cur = 1ll * x + 1ll * h[v] * k;
    cur %= maxc;
    IT1.up(1, 1, n, in[v], out[v], cur);
    cur = (-k + maxc) % maxc;
    IT2.up(1, 1, n, in[v], out[v], cur);
  } else {
    long long cur1 = IT1.get(1, 1, n, in[v]);
    long long cur2 = IT2.get(1, 1, n, in[v]);
    cout << (cur1 + cur2 * 1ll * h[v]) % maxc << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int u;
    cin >> u;
    adj[u].push_back(i);
  }
  dfs(1, -1);
  cin >> q;
  while (q--) query();
  return 0;
}

```