#include <bits/stdc++.h>
using namespace std;
inline int II() {
  int n;
  scanf("%d", &n);
  return n;
}
struct ii {
  long long v, k;
};
ii ft[300002];
int N;
void update(int k, long long deltav, long long deltak) {
  for (; k; k -= (k & (-k))) {
    ft[k].v = (ft[k].v + deltav + 1000000007LL) % 1000000007LL;
    ft[k].k = (ft[k].k + deltak + 1000000007LL) % 1000000007LL;
  }
}
long long sum_v(int k) {
  long long ans = 0;
  for (; k <= N; k += (k & (-k)))
    ans = (ans + ft[k].v + 1000000007LL) % 1000000007LL;
  return ans;
}
long long sum_k(int k) {
  long long ans = 0;
  for (; k <= N; k += (k & (-k)))
    ans = (ans + ft[k].k + 1000000007LL) % 1000000007LL;
  return ans;
}
int istart[300002], iend[300002];
int c = 1;
int father[300002];
int depth[300002];
vector<int> adj[300002];
void fatina_dfs(int u, int d) {
  depth[u] = d;
  istart[u] = c;
  c++;
  for (int v : adj[u]) fatina_dfs(v, d + 1);
  iend[u] = c - 1;
}
int main() {
  N = II();
  depth[1] = 0;
  for (int x = 2; x <= N; x++) {
    father[x] = II();
    adj[father[x]].push_back(x);
  }
  fatina_dfs(1, 0);
  int Q = II();
  while (Q--) {
    int t = II();
    if (t == 1) {
      long long v = II(), x = II(), k = II();
      x += (k * depth[v]) % 1000000007LL;
      x %= 1000000007LL;
      update(istart[v] - 1, -x, -k);
      update(iend[v], x, k);
    } else {
      int v = II();
      long long sum = sum_v(istart[v]) % 1000000007LL;
      long long k = sum_k(istart[v]) % 1000000007LL;
      long long ris = sum - (k * depth[v]);
      if (ris < 0) {
        long long temp = -ris;
        temp /= 1000000007LL;
        temp++;
        temp *= 1000000007LL;
        ris += temp;
      }
      ris %= 1000000007LL;
      printf("%d\n", (int)ris);
    }
  }
}

### Prompt

Please formulate a Cpp solution to 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;
inline int II() {
  int n;
  scanf("%d", &n);
  return n;
}
struct ii {
  long long v, k;
};
ii ft[300002];
int N;
void update(int k, long long deltav, long long deltak) {
  for (; k; k -= (k & (-k))) {
    ft[k].v = (ft[k].v + deltav + 1000000007LL) % 1000000007LL;
    ft[k].k = (ft[k].k + deltak + 1000000007LL) % 1000000007LL;
  }
}
long long sum_v(int k) {
  long long ans = 0;
  for (; k <= N; k += (k & (-k)))
    ans = (ans + ft[k].v + 1000000007LL) % 1000000007LL;
  return ans;
}
long long sum_k(int k) {
  long long ans = 0;
  for (; k <= N; k += (k & (-k)))
    ans = (ans + ft[k].k + 1000000007LL) % 1000000007LL;
  return ans;
}
int istart[300002], iend[300002];
int c = 1;
int father[300002];
int depth[300002];
vector<int> adj[300002];
void fatina_dfs(int u, int d) {
  depth[u] = d;
  istart[u] = c;
  c++;
  for (int v : adj[u]) fatina_dfs(v, d + 1);
  iend[u] = c - 1;
}
int main() {
  N = II();
  depth[1] = 0;
  for (int x = 2; x <= N; x++) {
    father[x] = II();
    adj[father[x]].push_back(x);
  }
  fatina_dfs(1, 0);
  int Q = II();
  while (Q--) {
    int t = II();
    if (t == 1) {
      long long v = II(), x = II(), k = II();
      x += (k * depth[v]) % 1000000007LL;
      x %= 1000000007LL;
      update(istart[v] - 1, -x, -k);
      update(iend[v], x, k);
    } else {
      int v = II();
      long long sum = sum_v(istart[v]) % 1000000007LL;
      long long k = sum_k(istart[v]) % 1000000007LL;
      long long ris = sum - (k * depth[v]);
      if (ris < 0) {
        long long temp = -ris;
        temp /= 1000000007LL;
        temp++;
        temp *= 1000000007LL;
        ris += temp;
      }
      ris %= 1000000007LL;
      printf("%d\n", (int)ris);
    }
  }
}

```