#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int MaxN = int(4e5) + 256;
const int MOD = int(1e9) + 7;
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
inline bool Palindrome(const string& s) {
  return equal(s.begin(), s.end(), s.rbegin());
}
int n;
vector<int> g[MaxN];
int tin[MaxN], tout[MaxN], timer, depth[MaxN];
void dfs(int v, int dep) {
  tin[v] = ++timer;
  depth[v] = dep;
  for (auto to : g[v]) dfs(to, dep + 1);
  tout[v] = timer;
}
class SegmentTree {
  struct node {
    long long val, p;
    node() { val = p = 0; }
  };

 public:
  vector<node> t;
  int n;
  SegmentTree(int n = 1) {
    int k = 1;
    while (k < n) k += k;
    this->n = k;
    t.resize(12 * k);
  }
  inline void push(int v, int sz) {
    if (t[v].p) {
      if (sz != 1) {
        t[v + v].p = (1ll * t[v + v].p + t[v].p) % MOD;
        t[v + v + 1].p = (1ll * t[v + v + 1].p + t[v].p) % MOD;
      }
      t[v].val = (t[v].val + ((sz * 1ll * t[v].p) % MOD) + MOD) % MOD;
      t[v].p = 0;
    }
  }
  void update(int v, int tl, int tr, int l, int r, int val) {
    if (l > tr || tl > r) return;
    push(v, tr - tl + 1);
    if (l <= tl && tr <= r) {
      t[v].p = (t[v].p + val) % MOD;
      push(v, tr - tl + 1);
      return;
    }
    int tm = (tl + tr) >> 1;
    update(v + v, tl, tm, l, r, val);
    update(v + v + 1, tm + 1, tr, l, r, val);
    t[v].val = (1ll * t[v + v].val + t[v + v + 1].val + MOD) % MOD;
  }
  int get(int v, int l, int r, int pos) {
    push(v, r - l + 1);
    if (l == r) return t[v].val;
    int m = (l + r) >> 1;
    if (pos <= m) return get(v + v, l, m, pos);
    return get(v + v + 1, m + 1, r, pos);
  }
  inline void update(int l, int r, int val) { update(1, 0, n, l, r, val); }
  inline int get(int pos) { return get(1, 0, n, pos); }
};
SegmentTree s1, s2;
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int par;
    scanf("%d", &par);
    g[par].push_back(i);
  }
  dfs(1, 1);
  s1 = SegmentTree(timer);
  s2 = SegmentTree(timer);
  int q;
  scanf("%d", &q);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int v, x, k;
      scanf("%d%d%d", &v, &x, &k);
      s1.update(tin[v], tout[v], (x + 1ll * depth[v] * k % MOD) % MOD);
      s2.update(tin[v], tout[v], k);
    } else {
      int v;
      scanf("%d", &v);
      int res = s1.get(tin[v]);
      res = (res - (1ll * depth[v] * s2.get(tin[v])) % MOD);
      res = (res + MOD) % MOD;
      printf("%d\n", res);
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
const double eps = 1e-9;
const int MaxN = int(4e5) + 256;
const int MOD = int(1e9) + 7;
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
inline bool Palindrome(const string& s) {
  return equal(s.begin(), s.end(), s.rbegin());
}
int n;
vector<int> g[MaxN];
int tin[MaxN], tout[MaxN], timer, depth[MaxN];
void dfs(int v, int dep) {
  tin[v] = ++timer;
  depth[v] = dep;
  for (auto to : g[v]) dfs(to, dep + 1);
  tout[v] = timer;
}
class SegmentTree {
  struct node {
    long long val, p;
    node() { val = p = 0; }
  };

 public:
  vector<node> t;
  int n;
  SegmentTree(int n = 1) {
    int k = 1;
    while (k < n) k += k;
    this->n = k;
    t.resize(12 * k);
  }
  inline void push(int v, int sz) {
    if (t[v].p) {
      if (sz != 1) {
        t[v + v].p = (1ll * t[v + v].p + t[v].p) % MOD;
        t[v + v + 1].p = (1ll * t[v + v + 1].p + t[v].p) % MOD;
      }
      t[v].val = (t[v].val + ((sz * 1ll * t[v].p) % MOD) + MOD) % MOD;
      t[v].p = 0;
    }
  }
  void update(int v, int tl, int tr, int l, int r, int val) {
    if (l > tr || tl > r) return;
    push(v, tr - tl + 1);
    if (l <= tl && tr <= r) {
      t[v].p = (t[v].p + val) % MOD;
      push(v, tr - tl + 1);
      return;
    }
    int tm = (tl + tr) >> 1;
    update(v + v, tl, tm, l, r, val);
    update(v + v + 1, tm + 1, tr, l, r, val);
    t[v].val = (1ll * t[v + v].val + t[v + v + 1].val + MOD) % MOD;
  }
  int get(int v, int l, int r, int pos) {
    push(v, r - l + 1);
    if (l == r) return t[v].val;
    int m = (l + r) >> 1;
    if (pos <= m) return get(v + v, l, m, pos);
    return get(v + v + 1, m + 1, r, pos);
  }
  inline void update(int l, int r, int val) { update(1, 0, n, l, r, val); }
  inline int get(int pos) { return get(1, 0, n, pos); }
};
SegmentTree s1, s2;
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int par;
    scanf("%d", &par);
    g[par].push_back(i);
  }
  dfs(1, 1);
  s1 = SegmentTree(timer);
  s2 = SegmentTree(timer);
  int q;
  scanf("%d", &q);
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int v, x, k;
      scanf("%d%d%d", &v, &x, &k);
      s1.update(tin[v], tout[v], (x + 1ll * depth[v] * k % MOD) % MOD);
      s2.update(tin[v], tout[v], k);
    } else {
      int v;
      scanf("%d", &v);
      int res = s1.get(tin[v]);
      res = (res - (1ll * depth[v] * s2.get(tin[v])) % MOD);
      res = (res + MOD) % MOD;
      printf("%d\n", res);
    }
  }
  return 0;
}

```