#include <bits/stdc++.h>
using namespace std;
const long long Mod = 1e9 + 7;
const int maxn = 300010;
int it[maxn * 4], lazy[maxn * 4];
int it2[maxn * 4], lazy2[maxn * 4];
int st[maxn], ed[maxn], ord;
int lvl[maxn];
int n;
vector<int> e[maxn];
void dfs(int u, int pa) {
  st[u] = ++ord;
  lvl[u] = lvl[pa] + 1;
  for (int i = (0), _b = (int((e[u]).size())); i < _b; ++i) {
    dfs(e[u][i], u);
  }
  ed[u] = ord;
}
void add(int &x, int y) {
  x += y;
  x %= Mod;
}
void upd(int k) {
  int lson = k << 1, rson = lson | 1;
  add(lazy[lson], lazy[k]);
  add(lazy[rson], lazy[k]);
  add(lazy2[lson], lazy2[k]);
  add(lazy2[rson], lazy2[k]);
  add(it[lson], lazy[k]);
  add(it[rson], lazy[k]);
  add(it2[lson], lazy2[k]);
  add(it2[rson], lazy2[k]);
  lazy[k] = lazy2[k] = 0;
}
void update(int k, int l, int r, int u, int v, int c, int c2) {
  if (l > v || r < u) return;
  if (u <= l && r <= v) {
    add(it[k], c);
    add(lazy[k], c);
    add(it2[k], c2);
    add(lazy2[k], c2);
    return;
  }
  upd(k);
  int mid = (l + r) >> 1, lson = k << 1, rson = lson | 1;
  update(lson, l, mid, u, v, c, c2);
  update(rson, mid + 1, r, u, v, c, c2);
}
int query(int k, int l, int r, int u, int lv) {
  if (u == l && u == r) {
    return (it[k] - 1ll * it2[k] * lv + Mod * Mod) % Mod;
  }
  upd(k);
  int mid = (l + r) >> 1, lson = k << 1, rson = lson | 1;
  int ans;
  if (u <= mid)
    return query(lson, l, mid, u, lv);
  else
    return query(rson, mid + 1, r, u, lv);
}
int main() {
  scanf("%d", &n);
  for (int i = (2), _b = (n); i <= _b; ++i) {
    int x;
    scanf("%d", &x);
    e[x].push_back(i);
  }
  dfs(1, 0);
  int q;
  scanf("%d", &q);
  while (q--) {
    int tag;
    scanf("%d", &tag);
    if (tag == 1) {
      int u, x, k;
      scanf("%d%d%d", &u, &x, &k);
      update(1, 1, n, st[u], ed[u], (x + 1ll * k * lvl[u]) % Mod, k);
    } else {
      int u;
      scanf("%d", &u);
      printf("%d\n", query(1, 1, n, st[u], lvl[u]));
    }
  }
  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 long long Mod = 1e9 + 7;
const int maxn = 300010;
int it[maxn * 4], lazy[maxn * 4];
int it2[maxn * 4], lazy2[maxn * 4];
int st[maxn], ed[maxn], ord;
int lvl[maxn];
int n;
vector<int> e[maxn];
void dfs(int u, int pa) {
  st[u] = ++ord;
  lvl[u] = lvl[pa] + 1;
  for (int i = (0), _b = (int((e[u]).size())); i < _b; ++i) {
    dfs(e[u][i], u);
  }
  ed[u] = ord;
}
void add(int &x, int y) {
  x += y;
  x %= Mod;
}
void upd(int k) {
  int lson = k << 1, rson = lson | 1;
  add(lazy[lson], lazy[k]);
  add(lazy[rson], lazy[k]);
  add(lazy2[lson], lazy2[k]);
  add(lazy2[rson], lazy2[k]);
  add(it[lson], lazy[k]);
  add(it[rson], lazy[k]);
  add(it2[lson], lazy2[k]);
  add(it2[rson], lazy2[k]);
  lazy[k] = lazy2[k] = 0;
}
void update(int k, int l, int r, int u, int v, int c, int c2) {
  if (l > v || r < u) return;
  if (u <= l && r <= v) {
    add(it[k], c);
    add(lazy[k], c);
    add(it2[k], c2);
    add(lazy2[k], c2);
    return;
  }
  upd(k);
  int mid = (l + r) >> 1, lson = k << 1, rson = lson | 1;
  update(lson, l, mid, u, v, c, c2);
  update(rson, mid + 1, r, u, v, c, c2);
}
int query(int k, int l, int r, int u, int lv) {
  if (u == l && u == r) {
    return (it[k] - 1ll * it2[k] * lv + Mod * Mod) % Mod;
  }
  upd(k);
  int mid = (l + r) >> 1, lson = k << 1, rson = lson | 1;
  int ans;
  if (u <= mid)
    return query(lson, l, mid, u, lv);
  else
    return query(rson, mid + 1, r, u, lv);
}
int main() {
  scanf("%d", &n);
  for (int i = (2), _b = (n); i <= _b; ++i) {
    int x;
    scanf("%d", &x);
    e[x].push_back(i);
  }
  dfs(1, 0);
  int q;
  scanf("%d", &q);
  while (q--) {
    int tag;
    scanf("%d", &tag);
    if (tag == 1) {
      int u, x, k;
      scanf("%d%d%d", &u, &x, &k);
      update(1, 1, n, st[u], ed[u], (x + 1ll * k * lvl[u]) % Mod, k);
    } else {
      int u;
      scanf("%d", &u);
      printf("%d\n", query(1, 1, n, st[u], lvl[u]));
    }
  }
  return 0;
}

```