#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
struct range {
  int l, r;
} R[400000];
vector<int> G[400000];
int n, q, root, stp, dep[400000];
void dfs(int u, int d) {
  R[u].l = ++stp;
  dep[u] = d;
  for (int i = 0; i < G[u].size(); i++) {
    dfs(G[u][i], d + 1);
  }
  R[u].r = stp;
}
long long T1[400000], T2[400000], T3[400000];
void modify(long long* T, int p, long long x) {
  x %= MOD;
  for (; p <= n; p += (p & (-p))) (T[p] += x) %= MOD;
}
long long query(long long* T, int p) {
  long long ans = 0;
  for (; p; p -= (p & (-p))) (ans += T[p]) %= MOD;
  return ans;
}
int main() {
  scanf("%d", &n);
  root = 1;
  for (int i = 2; i <= n; i++) {
    int t;
    scanf("%d", &t);
    if (t != i)
      G[t].push_back(i);
    else
      root = i;
  }
  dfs(root, 1);
  scanf("%d", &q);
  while (q--) {
    int typ, u, v, w;
    scanf("%d", &typ);
    if (typ == 1) {
      scanf("%d%d%d", &u, &v, &w);
      modify(T1, R[u].l, v);
      modify(T1, R[u].r + 1, MOD - v);
      modify(T2, R[u].l, w);
      modify(T2, R[u].r + 1, MOD - w);
      modify(T3, R[u].l, (long long)w * dep[u] % MOD);
      modify(T3, R[u].r + 1, MOD - (long long)w * dep[u] % MOD);
    } else {
      scanf("%d", &u);
      long long ans = query(T1, R[u].l);
      (ans -= query(T2, R[u].l) * (long long)dep[u] % MOD) %= MOD;
      ans += query(T3, R[u].l);
      ans %= MOD;
      if (ans < 0) ans += MOD;
      printf("%d\n", (int)ans);
    }
  }
}

### Prompt

Create a solution in Cpp for 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;
const long long MOD = 1000000007;
struct range {
  int l, r;
} R[400000];
vector<int> G[400000];
int n, q, root, stp, dep[400000];
void dfs(int u, int d) {
  R[u].l = ++stp;
  dep[u] = d;
  for (int i = 0; i < G[u].size(); i++) {
    dfs(G[u][i], d + 1);
  }
  R[u].r = stp;
}
long long T1[400000], T2[400000], T3[400000];
void modify(long long* T, int p, long long x) {
  x %= MOD;
  for (; p <= n; p += (p & (-p))) (T[p] += x) %= MOD;
}
long long query(long long* T, int p) {
  long long ans = 0;
  for (; p; p -= (p & (-p))) (ans += T[p]) %= MOD;
  return ans;
}
int main() {
  scanf("%d", &n);
  root = 1;
  for (int i = 2; i <= n; i++) {
    int t;
    scanf("%d", &t);
    if (t != i)
      G[t].push_back(i);
    else
      root = i;
  }
  dfs(root, 1);
  scanf("%d", &q);
  while (q--) {
    int typ, u, v, w;
    scanf("%d", &typ);
    if (typ == 1) {
      scanf("%d%d%d", &u, &v, &w);
      modify(T1, R[u].l, v);
      modify(T1, R[u].r + 1, MOD - v);
      modify(T2, R[u].l, w);
      modify(T2, R[u].r + 1, MOD - w);
      modify(T3, R[u].l, (long long)w * dep[u] % MOD);
      modify(T3, R[u].r + 1, MOD - (long long)w * dep[u] % MOD);
    } else {
      scanf("%d", &u);
      long long ans = query(T1, R[u].l);
      (ans -= query(T2, R[u].l) * (long long)dep[u] % MOD) %= MOD;
      ans += query(T3, R[u].l);
      ans %= MOD;
      if (ans < 0) ans += MOD;
      printf("%d\n", (int)ans);
    }
  }
}

```