#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const int maxn = 300001;
struct Fenwick {
  int n;
  long long s[600001];
  void add(int x, long long t) {
    while (x <= n) {
      s[x] = (s[x] + t) % mod;
      if (s[x] < 0) s[x] = (s[x] + mod) % mod;
      x += ((x) & (-(x)));
    }
  }
  int sum(int x) {
    long long ans = 0;
    while (x > 0) {
      ans = (ans + s[x]) % mod;
      x -= ((x) & (-(x)));
    }
    return (ans + mod) % mod;
  }
};
int n, m;
int h[maxn], ne[maxn], fa[maxn];
int dfn[maxn], st[maxn], ed[maxn];
long long dep[maxn];
Fenwick t1, t2;
void dfs(int x) {
  dep[x] = dep[fa[x]] + 1;
  dfn[++dfn[0]] = x;
  st[x] = dfn[0];
  for (int i = h[x]; i; i = ne[i]) dfs(i);
  ed[x] = dfn[0];
}
int main() {
  scanf("%d", &n);
  t1.n = t2.n = n;
  for (int i = 2; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    fa[i] = x;
    ne[i] = h[x];
    h[x] = i;
  }
  dfs(1);
  for (scanf("%d", &m); m > 0; --m) {
    int x;
    scanf("%d", &x);
    if (x == 1) {
      long long v, x, k;
      scanf("%I64d%I64d%I64d", &v, &x, &k);
      t1.add(st[v], (x + dep[v] * k) % mod);
      t1.add(ed[v] + 1, (-x - dep[v] * k) % mod);
      t2.add(st[v], k);
      t2.add(ed[v] + 1, -k);
    } else {
      scanf("%d", &x);
      printf("%d\n",
             ((t1.sum(st[x]) - t2.sum(st[x]) * dep[x]) % mod + mod) % mod);
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 = 1000000007;
const int maxn = 300001;
struct Fenwick {
  int n;
  long long s[600001];
  void add(int x, long long t) {
    while (x <= n) {
      s[x] = (s[x] + t) % mod;
      if (s[x] < 0) s[x] = (s[x] + mod) % mod;
      x += ((x) & (-(x)));
    }
  }
  int sum(int x) {
    long long ans = 0;
    while (x > 0) {
      ans = (ans + s[x]) % mod;
      x -= ((x) & (-(x)));
    }
    return (ans + mod) % mod;
  }
};
int n, m;
int h[maxn], ne[maxn], fa[maxn];
int dfn[maxn], st[maxn], ed[maxn];
long long dep[maxn];
Fenwick t1, t2;
void dfs(int x) {
  dep[x] = dep[fa[x]] + 1;
  dfn[++dfn[0]] = x;
  st[x] = dfn[0];
  for (int i = h[x]; i; i = ne[i]) dfs(i);
  ed[x] = dfn[0];
}
int main() {
  scanf("%d", &n);
  t1.n = t2.n = n;
  for (int i = 2; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    fa[i] = x;
    ne[i] = h[x];
    h[x] = i;
  }
  dfs(1);
  for (scanf("%d", &m); m > 0; --m) {
    int x;
    scanf("%d", &x);
    if (x == 1) {
      long long v, x, k;
      scanf("%I64d%I64d%I64d", &v, &x, &k);
      t1.add(st[v], (x + dep[v] * k) % mod);
      t1.add(ed[v] + 1, (-x - dep[v] * k) % mod);
      t2.add(st[v], k);
      t2.add(ed[v] + 1, -k);
    } else {
      scanf("%d", &x);
      printf("%d\n",
             ((t1.sum(st[x]) - t2.sum(st[x]) * dep[x]) % mod + mod) % mod);
    }
  }
  return 0;
}

```