#include <bits/stdc++.h>
using namespace std;
int n, timer;
vector<int> g[300005];
int a[2 * 300005], C1[4 * 2 * 300005], C2[4 * 2 * 300005], tin[300005],
    tout[300005], h[300005];
void dfs(int v, int hv, int p) {
  h[v] = hv;
  a[timer] = hv;
  tin[v] = timer++;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to != p) dfs(to, hv + 1, v);
  }
  tout[v] = timer++;
}
void push(int v, int tl, int tr) {
  if (tl != tr) {
    int v1 = 2 * v, v2 = 2 * v + 1;
    C1[v1] = (C1[v1] + C1[v]) % 1000000007;
    C1[v2] = (C1[v2] + C1[v]) % 1000000007;
    C2[v2] = (C2[v2] + C2[v]) % 1000000007;
    C2[v1] = (C2[v1] + C2[v]) % 1000000007;
    C1[v] = 0, C2[v] = 0;
  }
}
void update(int v, int tl, int tr, int l, int r, long long c1, long long c2) {
  push(v, tl, tr);
  if (l > r) return;
  if (tl == l && tr == r) {
    C1[v] = (C1[v] + c1) % 1000000007, C2[v] = (C2[v] + c2) % 1000000007;
    push(v, tl, tr);
    return;
  }
  int mid = (tl + tr) / 2, v1 = 2 * v, v2 = 2 * v + 1;
  update(v1, tl, mid, l, min(mid, r), c1, c2);
  update(v2, mid + 1, tr, max(mid + 1, l), r, c1, c2);
}
long long query(int v, int tl, int tr, int pos) {
  push(v, tl, tr);
  if (tl == tr)
    return ((C1[v] - ((long long)a[tl] * (long long)C2[v]) % 1000000007) %
                1000000007 +
            1000000007) %
           1000000007;
  int mid = (tl + tr) / 2, v1 = 2 * v, v2 = 2 * v + 1;
  if (pos <= mid) return query(v1, tl, mid, pos);
  return query(v2, mid + 1, tr, pos);
}
int main() {
  cin >> n;
  timer = 1;
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    g[x].push_back(i);
  }
  dfs(1, 0, 0);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int tip;
    scanf("%d", &tip);
    if (tip == 1) {
      int v, x, k;
      scanf("%d %d %d", &v, &x, &k);
      long long c1 =
          (((long long)k * (long long)h[v]) % 1000000007 + x) % 1000000007;
      long long c2 = k % 1000000007;
      update(1, 1, timer, tin[v], tout[v], c1, c2);
    } else {
      int v;
      scanf("%d", &v);
      printf("%d\n", (int)query(1, 1, timer, tin[v]));
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int n, timer;
vector<int> g[300005];
int a[2 * 300005], C1[4 * 2 * 300005], C2[4 * 2 * 300005], tin[300005],
    tout[300005], h[300005];
void dfs(int v, int hv, int p) {
  h[v] = hv;
  a[timer] = hv;
  tin[v] = timer++;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to != p) dfs(to, hv + 1, v);
  }
  tout[v] = timer++;
}
void push(int v, int tl, int tr) {
  if (tl != tr) {
    int v1 = 2 * v, v2 = 2 * v + 1;
    C1[v1] = (C1[v1] + C1[v]) % 1000000007;
    C1[v2] = (C1[v2] + C1[v]) % 1000000007;
    C2[v2] = (C2[v2] + C2[v]) % 1000000007;
    C2[v1] = (C2[v1] + C2[v]) % 1000000007;
    C1[v] = 0, C2[v] = 0;
  }
}
void update(int v, int tl, int tr, int l, int r, long long c1, long long c2) {
  push(v, tl, tr);
  if (l > r) return;
  if (tl == l && tr == r) {
    C1[v] = (C1[v] + c1) % 1000000007, C2[v] = (C2[v] + c2) % 1000000007;
    push(v, tl, tr);
    return;
  }
  int mid = (tl + tr) / 2, v1 = 2 * v, v2 = 2 * v + 1;
  update(v1, tl, mid, l, min(mid, r), c1, c2);
  update(v2, mid + 1, tr, max(mid + 1, l), r, c1, c2);
}
long long query(int v, int tl, int tr, int pos) {
  push(v, tl, tr);
  if (tl == tr)
    return ((C1[v] - ((long long)a[tl] * (long long)C2[v]) % 1000000007) %
                1000000007 +
            1000000007) %
           1000000007;
  int mid = (tl + tr) / 2, v1 = 2 * v, v2 = 2 * v + 1;
  if (pos <= mid) return query(v1, tl, mid, pos);
  return query(v2, mid + 1, tr, pos);
}
int main() {
  cin >> n;
  timer = 1;
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    g[x].push_back(i);
  }
  dfs(1, 0, 0);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int tip;
    scanf("%d", &tip);
    if (tip == 1) {
      int v, x, k;
      scanf("%d %d %d", &v, &x, &k);
      long long c1 =
          (((long long)k * (long long)h[v]) % 1000000007 + x) % 1000000007;
      long long c2 = k % 1000000007;
      update(1, 1, timer, tin[v], tout[v], c1, c2);
    } else {
      int v;
      scanf("%d", &v);
      printf("%d\n", (int)query(1, 1, timer, tin[v]));
    }
  }
}

```