#include <bits/stdc++.h>
using namespace std;
long long sum[333333], sk[333333];
int deg[333333], lo[333333], hi[333333];
char str[333333];
vector<int> G[333333];
int dis, n;
void dfs(int u, int d) {
  lo[u] = ++dis;
  deg[u] = d;
  for (int i = 0; i < G[u].size(); i++) {
    dfs(G[u][i], d + 1);
  }
  hi[u] = dis;
}
int lowbit(int x) { return x & -x; }
void add(long long *c, int x, long long d) {
  while (x <= n) {
    c[x] = (c[x] + d) % 1000000007;
    x += lowbit(x);
  }
}
long long query(long long *c, int x) {
  long long ret = 0;
  while (x > 0) {
    ret = (ret + c[x]) % 1000000007;
    x -= lowbit(x);
  }
  return ret;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(sum, 0, sizeof(sum));
    memset(sk, 0, sizeof(sk));
    memset(deg, 0, sizeof(deg));
    for (int i = 1; i <= n; i++) G[i].clear();
    for (int i = 2; i <= n; i++) {
      int fa;
      scanf("%d", &fa);
      G[fa].push_back(i);
    }
    dis = 0;
    dfs(1, 0);
    int m;
    scanf("%d", &m);
    while (m--) {
      int op, v;
      long long k, x;
      scanf("%d", &op);
      if (op == 1) {
        scanf("%d%I64d%I64d", &v, &x, &k);
        add(sum, lo[v], x + k * deg[v]);
        add(sum, hi[v] + 1, -(x + k * deg[v]));
        add(sk, lo[v], k);
        add(sk, hi[v] + 1, -k);
      } else {
        scanf("%d", &v);
        long long ans = query(sum, lo[v]) - query(sk, lo[v]) * deg[v];
        printf("%I64d\n", (ans % 1000000007 + 1000000007) % 1000000007);
      }
    }
  }
  return 0;
}

### Prompt

Generate 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;
long long sum[333333], sk[333333];
int deg[333333], lo[333333], hi[333333];
char str[333333];
vector<int> G[333333];
int dis, n;
void dfs(int u, int d) {
  lo[u] = ++dis;
  deg[u] = d;
  for (int i = 0; i < G[u].size(); i++) {
    dfs(G[u][i], d + 1);
  }
  hi[u] = dis;
}
int lowbit(int x) { return x & -x; }
void add(long long *c, int x, long long d) {
  while (x <= n) {
    c[x] = (c[x] + d) % 1000000007;
    x += lowbit(x);
  }
}
long long query(long long *c, int x) {
  long long ret = 0;
  while (x > 0) {
    ret = (ret + c[x]) % 1000000007;
    x -= lowbit(x);
  }
  return ret;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(sum, 0, sizeof(sum));
    memset(sk, 0, sizeof(sk));
    memset(deg, 0, sizeof(deg));
    for (int i = 1; i <= n; i++) G[i].clear();
    for (int i = 2; i <= n; i++) {
      int fa;
      scanf("%d", &fa);
      G[fa].push_back(i);
    }
    dis = 0;
    dfs(1, 0);
    int m;
    scanf("%d", &m);
    while (m--) {
      int op, v;
      long long k, x;
      scanf("%d", &op);
      if (op == 1) {
        scanf("%d%I64d%I64d", &v, &x, &k);
        add(sum, lo[v], x + k * deg[v]);
        add(sum, hi[v] + 1, -(x + k * deg[v]));
        add(sk, lo[v], k);
        add(sk, hi[v] + 1, -k);
      } else {
        scanf("%d", &v);
        long long ans = query(sum, lo[v]) - query(sk, lo[v]) * deg[v];
        printf("%I64d\n", (ans % 1000000007 + 1000000007) % 1000000007);
      }
    }
  }
  return 0;
}

```