#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
class BIT {
  long long MX, *bit;

 public:
  BIT() {
    MX = 0;
    bit = NULL;
  };
  void init(int N) {
    MX = N;
    bit = new long long[MX + 1];
    for (int j = 0; j <= MX; j++) bit[j] = 0;
  }
  int get(int x) {
    long long ret = 0;
    while (x > 0) {
      ret += bit[x];
      ret += MOD;
      ret %= MOD;
      x -= x & (-x);
    }
    return ret;
  }
  void add(int x, int V) {
    while (x <= MX) {
      bit[x] += V;
      bit[x] += MOD;
      bit[x] %= MOD;
      x += x & (-x);
    }
  }
} tree1, tree2;
const int MX = (1 << 20);
int in[MX], out[MX], n, timer, depth[MX];
vector<int> v[MX];
void dfs(int x) {
  in[x] = ++timer;
  for (auto nxt : v[x]) {
    depth[nxt] = depth[x] + 1;
    dfs(nxt);
  }
  out[x] = timer;
}
int main() {
  scanf("%d", &n);
  for (int j = 2; j <= n; j++) {
    int x;
    scanf("%d", &x);
    v[x].push_back(j);
  }
  dfs(1);
  int QN;
  scanf("%d", &QN);
  tree1.init(n);
  tree2.init(n);
  while (QN--) {
    int cmd, node;
    scanf("%d %d", &cmd, &node);
    if (cmd == 2) {
      long long ans = tree2.get(in[node]);
      ans += MOD;
      ans %= MOD;
      ans *= 1ll * depth[node];
      ans %= MOD;
      long long theta = tree1.get(in[node]);
      swap(ans, theta);
      ans -= theta;
      while (ans < 0) ans += MOD;
      ans %= MOD;
      printf("%lld\n", ans);
    } else {
      int X, K;
      scanf("%d %d", &X, &K);
      tree2.add(in[node], K);
      tree2.add(out[node] + 1, -K);
      long long theta = X + 1ll * depth[node] * K;
      theta %= MOD;
      tree1.add(in[node], theta);
      tree1.add(out[node] + 1, -theta);
    }
  }
}

### 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 int MOD = 1e9 + 7;
class BIT {
  long long MX, *bit;

 public:
  BIT() {
    MX = 0;
    bit = NULL;
  };
  void init(int N) {
    MX = N;
    bit = new long long[MX + 1];
    for (int j = 0; j <= MX; j++) bit[j] = 0;
  }
  int get(int x) {
    long long ret = 0;
    while (x > 0) {
      ret += bit[x];
      ret += MOD;
      ret %= MOD;
      x -= x & (-x);
    }
    return ret;
  }
  void add(int x, int V) {
    while (x <= MX) {
      bit[x] += V;
      bit[x] += MOD;
      bit[x] %= MOD;
      x += x & (-x);
    }
  }
} tree1, tree2;
const int MX = (1 << 20);
int in[MX], out[MX], n, timer, depth[MX];
vector<int> v[MX];
void dfs(int x) {
  in[x] = ++timer;
  for (auto nxt : v[x]) {
    depth[nxt] = depth[x] + 1;
    dfs(nxt);
  }
  out[x] = timer;
}
int main() {
  scanf("%d", &n);
  for (int j = 2; j <= n; j++) {
    int x;
    scanf("%d", &x);
    v[x].push_back(j);
  }
  dfs(1);
  int QN;
  scanf("%d", &QN);
  tree1.init(n);
  tree2.init(n);
  while (QN--) {
    int cmd, node;
    scanf("%d %d", &cmd, &node);
    if (cmd == 2) {
      long long ans = tree2.get(in[node]);
      ans += MOD;
      ans %= MOD;
      ans *= 1ll * depth[node];
      ans %= MOD;
      long long theta = tree1.get(in[node]);
      swap(ans, theta);
      ans -= theta;
      while (ans < 0) ans += MOD;
      ans %= MOD;
      printf("%lld\n", ans);
    } else {
      int X, K;
      scanf("%d %d", &X, &K);
      tree2.add(in[node], K);
      tree2.add(out[node] + 1, -K);
      long long theta = X + 1ll * depth[node] * K;
      theta %= MOD;
      tree1.add(in[node], theta);
      tree1.add(out[node] + 1, -theta);
    }
  }
}

```