#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
struct segtr {
  vector<long long> tr;
  long long nn;
  void build(long long tn) {
    tr.resize(2 * tn, 0);
    nn = tn;
  }
  void mix(long long l, long long r, long long val) {
    for (l += nn, r += nn; l < r; l >>= 1, r >>= 1) {
      if (l & 1) tr[l] += val, tr[l++] %= mod;
      if (r & 1) tr[--r] += val, tr[r] %= mod;
    }
  }
  long long get(long long p) {
    long long ans = 0;
    for (p += nn; p >= 1; p >>= 1) ans += tr[p], ans %= mod;
    return ans;
  }
};
segtr su, rates;
long long cnt = 0, dep = 0, n, m, ans[524288], sans[524288];
vector<long long> que[524288], edg[524288];
vector<pair<long long, long long> > updrate[524288], updsu[524288];
void dfs(long long u) {
  for (auto&& q : updsu[u]) su.mix(q.second, m, q.first);
  for (auto&& q : updrate[u])
    rates.mix(q.second, m, q.first), su.mix(q.second, m, -dep * q.first);
  for (auto&& q : que[u])
    ans[q] = (((long long)rates.get(q) % mod) * (dep % mod) +
              ((long long)su.get(q) % mod)) %
             mod;
  dep++;
  for (auto&& v : edg[u]) dfs(v);
  dep--;
  for (auto&& q : updrate[u])
    rates.mix(q.second, m, -q.first), su.mix(q.second, m, dep * q.first);
  for (auto&& q : updsu[u]) su.mix(q.second, m, -q.first);
}
int main() {
  cin >> n;
  for (long long i = 2; i < n + 1; i++) {
    int p;
    scanf("%d", &p);
    edg[p].push_back(i);
  }
  cin >> m;
  su.build(m), rates.build(m);
  for (long long i = 0; i < m; i++) {
    int t;
    scanf("%d", &t);
    if (t == 2) {
      int a;
      scanf("%d", &a);
      que[a].push_back(i);
      sans[i] = 1;
    } else if (t == 1) {
      long long v, x, k;
      scanf("%lld%lld%lld", &v, &x, &k);
      updrate[v].push_back(make_pair(-k, i));
      updsu[v].push_back(make_pair(x, i));
    }
  }
  dfs(1);
  for (long long i = 0; i < m; i++)
    if (sans[i]) printf("%lld\n", (ans[i] + mod) % mod);
}

### 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;
long long mod = 1000000007;
struct segtr {
  vector<long long> tr;
  long long nn;
  void build(long long tn) {
    tr.resize(2 * tn, 0);
    nn = tn;
  }
  void mix(long long l, long long r, long long val) {
    for (l += nn, r += nn; l < r; l >>= 1, r >>= 1) {
      if (l & 1) tr[l] += val, tr[l++] %= mod;
      if (r & 1) tr[--r] += val, tr[r] %= mod;
    }
  }
  long long get(long long p) {
    long long ans = 0;
    for (p += nn; p >= 1; p >>= 1) ans += tr[p], ans %= mod;
    return ans;
  }
};
segtr su, rates;
long long cnt = 0, dep = 0, n, m, ans[524288], sans[524288];
vector<long long> que[524288], edg[524288];
vector<pair<long long, long long> > updrate[524288], updsu[524288];
void dfs(long long u) {
  for (auto&& q : updsu[u]) su.mix(q.second, m, q.first);
  for (auto&& q : updrate[u])
    rates.mix(q.second, m, q.first), su.mix(q.second, m, -dep * q.first);
  for (auto&& q : que[u])
    ans[q] = (((long long)rates.get(q) % mod) * (dep % mod) +
              ((long long)su.get(q) % mod)) %
             mod;
  dep++;
  for (auto&& v : edg[u]) dfs(v);
  dep--;
  for (auto&& q : updrate[u])
    rates.mix(q.second, m, -q.first), su.mix(q.second, m, dep * q.first);
  for (auto&& q : updsu[u]) su.mix(q.second, m, -q.first);
}
int main() {
  cin >> n;
  for (long long i = 2; i < n + 1; i++) {
    int p;
    scanf("%d", &p);
    edg[p].push_back(i);
  }
  cin >> m;
  su.build(m), rates.build(m);
  for (long long i = 0; i < m; i++) {
    int t;
    scanf("%d", &t);
    if (t == 2) {
      int a;
      scanf("%d", &a);
      que[a].push_back(i);
      sans[i] = 1;
    } else if (t == 1) {
      long long v, x, k;
      scanf("%lld%lld%lld", &v, &x, &k);
      updrate[v].push_back(make_pair(-k, i));
      updsu[v].push_back(make_pair(x, i));
    }
  }
  dfs(1);
  for (long long i = 0; i < m; i++)
    if (sans[i]) printf("%lld\n", (ans[i] + mod) % mod);
}

```