#include <bits/stdc++.h>
using namespace std;
const int N = 3 * 1e5 + 10;
const long long mod = 1e9 + 7;
int n;
int m;
struct data {
  long long x;
  long long k;
  int t;
};
vector<data> ve[N];
vector<int> qr[N];
struct treearray {
  long long ta[N];
  inline void c(int x, long long t) {
    for (; x <= m; x += x & (-x)) (ta[x] += t) %= mod;
  }
  inline long long q(int x) {
    long long r = 0;
    for (; x; x -= x & (-x)) (r += ta[x]) %= mod;
    return r;
  }
} ta1, ta2;
int v[N];
int x[N];
int ct;
int al[N];
long long ans[N];
inline void add(int u, int V) {
  v[++ct] = V;
  x[ct] = al[u];
  al[u] = ct;
}
inline void dfs(int u, int dep) {
  for (vector<data>::iterator it = ve[u].begin(); it != ve[u].end(); ++it)
    ta1.c(it->t, (it->x + dep * it->k) % mod), ta2.c(it->t, it->k);
  for (vector<int>::iterator it = qr[u].begin(); it != qr[u].end(); ++it)
    ans[*it] = (ta1.q(*it) + mod - ta2.q(*it) * dep % mod) % mod;
  for (int i = al[u]; i; i = x[i]) dfs(v[i], dep + 1);
  for (vector<data>::iterator it = ve[u].begin(); it != ve[u].end(); ++it)
    ta1.c(it->t, mod - (it->x + dep * it->k) % mod), ta2.c(it->t, mod - it->k);
}
int main() {
  scanf("%d", &n);
  for (int i = 2, f; i <= n; i++) scanf("%d", &f), add(f, i);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) ans[i] = -1;
  for (int i = 1, t, u, x, k; i <= m; i++) {
    scanf("%d%d", &t, &u);
    if (t == 1)
      scanf("%d%d", &x, &k), ve[u].push_back((data){x, k, i});
    else
      qr[u].push_back(i);
  }
  dfs(1, 0);
  for (int i = 1; i <= m; i++)
    if (ans[i] != -1) printf("%I64d\n", ans[i]);
  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 int N = 3 * 1e5 + 10;
const long long mod = 1e9 + 7;
int n;
int m;
struct data {
  long long x;
  long long k;
  int t;
};
vector<data> ve[N];
vector<int> qr[N];
struct treearray {
  long long ta[N];
  inline void c(int x, long long t) {
    for (; x <= m; x += x & (-x)) (ta[x] += t) %= mod;
  }
  inline long long q(int x) {
    long long r = 0;
    for (; x; x -= x & (-x)) (r += ta[x]) %= mod;
    return r;
  }
} ta1, ta2;
int v[N];
int x[N];
int ct;
int al[N];
long long ans[N];
inline void add(int u, int V) {
  v[++ct] = V;
  x[ct] = al[u];
  al[u] = ct;
}
inline void dfs(int u, int dep) {
  for (vector<data>::iterator it = ve[u].begin(); it != ve[u].end(); ++it)
    ta1.c(it->t, (it->x + dep * it->k) % mod), ta2.c(it->t, it->k);
  for (vector<int>::iterator it = qr[u].begin(); it != qr[u].end(); ++it)
    ans[*it] = (ta1.q(*it) + mod - ta2.q(*it) * dep % mod) % mod;
  for (int i = al[u]; i; i = x[i]) dfs(v[i], dep + 1);
  for (vector<data>::iterator it = ve[u].begin(); it != ve[u].end(); ++it)
    ta1.c(it->t, mod - (it->x + dep * it->k) % mod), ta2.c(it->t, mod - it->k);
}
int main() {
  scanf("%d", &n);
  for (int i = 2, f; i <= n; i++) scanf("%d", &f), add(f, i);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) ans[i] = -1;
  for (int i = 1, t, u, x, k; i <= m; i++) {
    scanf("%d%d", &t, &u);
    if (t == 1)
      scanf("%d%d", &x, &k), ve[u].push_back((data){x, k, i});
    else
      qr[u].push_back(i);
  }
  dfs(1, 0);
  for (int i = 1; i <= m; i++)
    if (ans[i] != -1) printf("%I64d\n", ans[i]);
  return 0;
}

```