#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x * 10 + ch - '0');
    ch = getchar();
  }
  return x;
}
long long mod = 1e9 + 7;
vector<long long> G[300005];
long long f[300005], dep[300005], siz[300005], son[300005];
void dfs1(long long x, long long fa) {
  siz[x] = 1;
  long long _max = 0;
  for (long long i = 0; i < G[x].size(); i++) {
    long long y = G[x][i];
    if (y == fa) continue;
    dep[y] = dep[x] + 1;
    f[y] = x;
    dfs1(y, x);
    if (siz[y] > _max) {
      _max = siz[y];
      son[x] = y;
    }
    siz[x] += siz[y];
  }
}
long long cnt = 0;
long long seg[300005], pos[300005], top[300005];
void dfs2(long long x) {
  if (son[f[x]] == x)
    top[x] = top[f[x]];
  else
    top[x] = x;
  seg[++cnt] = x;
  pos[x] = cnt;
  if (son[x]) dfs2(son[x]);
  for (long long i = 0; i < G[x].size(); i++) {
    long long y = G[x][i];
    if (top[y] == 0) {
      dfs2(y);
    }
  }
}
struct Seg_Tree {
  struct Tree {
    long long l, r, x, tag;
  } t[2000005];
  void build(long long id, long long l, long long r) {
    t[id].l = l, t[id].r = r;
    if (l == r) {
      return;
    }
    long long mid = (l + r) / 2;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
  }
  void push_tag(long long id) {
    if (t[id].tag) {
      t[id << 1].tag = (t[id << 1].tag + t[id].tag) % mod;
      t[id << 1 | 1].tag = (t[id << 1 | 1].tag + t[id].tag) % mod;
      t[id << 1].x = (t[id << 1].x + t[id].tag) % mod;
      t[id << 1 | 1].x = (t[id << 1 | 1].x + t[id].tag) % mod;
      t[id].tag = 0;
    }
  }
  void push_down(long long id, long long l, long long r, long long x) {
    if (l <= t[id].l && t[id].r <= r) {
      t[id].x = (t[id].x + x) % mod;
      t[id].tag = (t[id].tag + x) % mod;
      return;
    }
    push_tag(id);
    long long mid = (t[id].l + t[id].r) / 2;
    if (l <= mid) push_down(id << 1, l, r, x);
    if (r > mid) push_down(id << 1 | 1, l, r, x);
  }
  long long get_ans(long long id, long long x) {
    if (t[id].l == t[id].r) return t[id].x;
    push_tag(id);
    long long mid = (t[id].l + t[id].r) / 2;
    if (x <= mid)
      return get_ans(id << 1, x);
    else
      return get_ans(id << 1 | 1, x);
  }
} A, B;
signed main() {
  long long n = read();
  for (long long i = 2; i <= n; i++) {
    long long x = read();
    G[i].push_back(x);
    G[x].push_back(i);
  }
  dfs1(1, 0);
  dfs2(1);
  A.build(1, 1, n);
  B.build(1, 1, n);
  long long Q = read();
  while (Q--) {
    long long op = read();
    if (op == 1) {
      long long v = read(), x = read(), k = read();
      A.push_down(1, pos[v], pos[v] + siz[v] - 1, (x + dep[v] * k) % mod);
      B.push_down(1, pos[v], pos[v] + siz[v] - 1, k);
    } else {
      long long v = read();
      long long ans =
          (A.get_ans(1, pos[v]) - B.get_ans(1, pos[v]) * dep[v] % mod + mod) %
          mod;
      printf("%lld\n", ans);
    }
  }
  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;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x * 10 + ch - '0');
    ch = getchar();
  }
  return x;
}
long long mod = 1e9 + 7;
vector<long long> G[300005];
long long f[300005], dep[300005], siz[300005], son[300005];
void dfs1(long long x, long long fa) {
  siz[x] = 1;
  long long _max = 0;
  for (long long i = 0; i < G[x].size(); i++) {
    long long y = G[x][i];
    if (y == fa) continue;
    dep[y] = dep[x] + 1;
    f[y] = x;
    dfs1(y, x);
    if (siz[y] > _max) {
      _max = siz[y];
      son[x] = y;
    }
    siz[x] += siz[y];
  }
}
long long cnt = 0;
long long seg[300005], pos[300005], top[300005];
void dfs2(long long x) {
  if (son[f[x]] == x)
    top[x] = top[f[x]];
  else
    top[x] = x;
  seg[++cnt] = x;
  pos[x] = cnt;
  if (son[x]) dfs2(son[x]);
  for (long long i = 0; i < G[x].size(); i++) {
    long long y = G[x][i];
    if (top[y] == 0) {
      dfs2(y);
    }
  }
}
struct Seg_Tree {
  struct Tree {
    long long l, r, x, tag;
  } t[2000005];
  void build(long long id, long long l, long long r) {
    t[id].l = l, t[id].r = r;
    if (l == r) {
      return;
    }
    long long mid = (l + r) / 2;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
  }
  void push_tag(long long id) {
    if (t[id].tag) {
      t[id << 1].tag = (t[id << 1].tag + t[id].tag) % mod;
      t[id << 1 | 1].tag = (t[id << 1 | 1].tag + t[id].tag) % mod;
      t[id << 1].x = (t[id << 1].x + t[id].tag) % mod;
      t[id << 1 | 1].x = (t[id << 1 | 1].x + t[id].tag) % mod;
      t[id].tag = 0;
    }
  }
  void push_down(long long id, long long l, long long r, long long x) {
    if (l <= t[id].l && t[id].r <= r) {
      t[id].x = (t[id].x + x) % mod;
      t[id].tag = (t[id].tag + x) % mod;
      return;
    }
    push_tag(id);
    long long mid = (t[id].l + t[id].r) / 2;
    if (l <= mid) push_down(id << 1, l, r, x);
    if (r > mid) push_down(id << 1 | 1, l, r, x);
  }
  long long get_ans(long long id, long long x) {
    if (t[id].l == t[id].r) return t[id].x;
    push_tag(id);
    long long mid = (t[id].l + t[id].r) / 2;
    if (x <= mid)
      return get_ans(id << 1, x);
    else
      return get_ans(id << 1 | 1, x);
  }
} A, B;
signed main() {
  long long n = read();
  for (long long i = 2; i <= n; i++) {
    long long x = read();
    G[i].push_back(x);
    G[x].push_back(i);
  }
  dfs1(1, 0);
  dfs2(1);
  A.build(1, 1, n);
  B.build(1, 1, n);
  long long Q = read();
  while (Q--) {
    long long op = read();
    if (op == 1) {
      long long v = read(), x = read(), k = read();
      A.push_down(1, pos[v], pos[v] + siz[v] - 1, (x + dep[v] * k) % mod);
      B.push_down(1, pos[v], pos[v] + siz[v] - 1, k);
    } else {
      long long v = read();
      long long ans =
          (A.get_ans(1, pos[v]) - B.get_ans(1, pos[v]) * dep[v] % mod + mod) %
          mod;
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```