#include <bits/stdc++.h>
using namespace std;
long long exp(long long b, long long e) {
  if (b == 1 || e == 0) return 1;
  if (e == 1) return b % 1000000007ll;
  if (e == 2) return (b * b) % 1000000007ll;
  return ((e % 2 == 0)
              ? exp(exp(b, e / 2), 2) % 1000000007ll
              : (b * (exp(exp(b, e / 2), 2) % 1000000007ll)) % 1000000007ll);
}
long long gcd(long long a, long long b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
pair<long long, long long> change[1100000], range[1100000 / 3];
int h[1100000 / 3];
class SegmentTree {
  vector<int> A, st;
  int n;
  int parent(int p) { return (p >> 1); }
  int left(int p) { return (p << 1); }
  int right(int p) { return (p << 1) + 1; }
  void update(int p, int L, int R, int i, int j, pair<long long, long long> val,
              int d) {
    if (L > j || R < i) return;
    if (L == i && R == j) {
      change[p].first += (val.first + d * val.second) % 1000000007ll;
      change[p].first %= 1000000007ll;
      change[p].second += val.second;
      change[p].second %= 1000000007ll;
      return;
    }
    int mid = (L + R) / 2;
    if (mid >= j) {
      update(left(p), L, mid, i, j, val, d);
      return;
    }
    if (mid < i) {
      update(right(p), mid + 1, R, i, j, val, d);
      return;
    }
    update(left(p), L, mid, i, mid, val, d);
    update(right(p), mid + 1, R, mid + 1, j, val, d);
  }
  long long query(int p, int L, int R, int pos, long long d,
                  pair<long long, long long> ans) {
    if (L == R) {
      return ((ans.first + change[p].first) % 1000000007ll -
              d * (ans.second + change[p].second) % 1000000007ll +
              1000000007ll) %
             1000000007ll;
    }
    int mid = (L + R) / 2;
    ans.first = (ans.first + change[p].first) % 1000000007ll;
    ans.second = (ans.second + change[p].second) % 1000000007ll;
    if (mid < pos) return query(right(p), mid + 1, R, pos, d, ans);
    return query(left(p), L, mid, pos, d, ans);
  }
  void build_st(int p, int L, int R) {
    change[p] = make_pair(0, 0);
    if (L == R) {
      return;
    }
    build_st(left(p), L, (L + R) / 2);
    build_st(right(p), (L + R) / 2 + 1, R);
  }

 public:
  SegmentTree(vector<int> &hol) {
    A = hol;
    n = A.size();
    build_st(1, 0, n - 1);
  }
  void update(int v, int x, int k) {
    update(1, 0, n - 1, range[v].first, range[v].second, make_pair(x, k), h[v]);
  }
  long long query(int v) {
    pair<long long, long long> ans = make_pair(0, 0);
    return query(1, 0, n - 1, range[v].first, h[v], ans);
  }
};
vector<int> a, ch[1100000 / 3];
int tm;
void dfs(int v, int p) {
  a.push_back(v);
  h[v] = h[p] + 1;
  range[v].first = tm++;
  for (vector<int>::iterator it = ch[v].begin(), itE = ch[v].end(); it != itE;
       ++it)
    dfs(*it, v);
  range[v].second = tm - 1;
}
int main() {
  h[0] = -1;
  tm = 0;
  int n, i, j, k;
  scanf("%d", &n);
  for (i = 2; i <= n; ++i) {
    scanf("%d", &j);
    ch[j].push_back(i);
  }
  dfs(1, 0);
  SegmentTree st(a);
  int q, v, x, t;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d %d %d", &v, &x, &k);
      st.update(v, x, k);
    } else {
      scanf("%d", &v);
      printf("%lld\n", (st.query(v) + 1000000007ll) % 1000000007ll);
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 exp(long long b, long long e) {
  if (b == 1 || e == 0) return 1;
  if (e == 1) return b % 1000000007ll;
  if (e == 2) return (b * b) % 1000000007ll;
  return ((e % 2 == 0)
              ? exp(exp(b, e / 2), 2) % 1000000007ll
              : (b * (exp(exp(b, e / 2), 2) % 1000000007ll)) % 1000000007ll);
}
long long gcd(long long a, long long b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
pair<long long, long long> change[1100000], range[1100000 / 3];
int h[1100000 / 3];
class SegmentTree {
  vector<int> A, st;
  int n;
  int parent(int p) { return (p >> 1); }
  int left(int p) { return (p << 1); }
  int right(int p) { return (p << 1) + 1; }
  void update(int p, int L, int R, int i, int j, pair<long long, long long> val,
              int d) {
    if (L > j || R < i) return;
    if (L == i && R == j) {
      change[p].first += (val.first + d * val.second) % 1000000007ll;
      change[p].first %= 1000000007ll;
      change[p].second += val.second;
      change[p].second %= 1000000007ll;
      return;
    }
    int mid = (L + R) / 2;
    if (mid >= j) {
      update(left(p), L, mid, i, j, val, d);
      return;
    }
    if (mid < i) {
      update(right(p), mid + 1, R, i, j, val, d);
      return;
    }
    update(left(p), L, mid, i, mid, val, d);
    update(right(p), mid + 1, R, mid + 1, j, val, d);
  }
  long long query(int p, int L, int R, int pos, long long d,
                  pair<long long, long long> ans) {
    if (L == R) {
      return ((ans.first + change[p].first) % 1000000007ll -
              d * (ans.second + change[p].second) % 1000000007ll +
              1000000007ll) %
             1000000007ll;
    }
    int mid = (L + R) / 2;
    ans.first = (ans.first + change[p].first) % 1000000007ll;
    ans.second = (ans.second + change[p].second) % 1000000007ll;
    if (mid < pos) return query(right(p), mid + 1, R, pos, d, ans);
    return query(left(p), L, mid, pos, d, ans);
  }
  void build_st(int p, int L, int R) {
    change[p] = make_pair(0, 0);
    if (L == R) {
      return;
    }
    build_st(left(p), L, (L + R) / 2);
    build_st(right(p), (L + R) / 2 + 1, R);
  }

 public:
  SegmentTree(vector<int> &hol) {
    A = hol;
    n = A.size();
    build_st(1, 0, n - 1);
  }
  void update(int v, int x, int k) {
    update(1, 0, n - 1, range[v].first, range[v].second, make_pair(x, k), h[v]);
  }
  long long query(int v) {
    pair<long long, long long> ans = make_pair(0, 0);
    return query(1, 0, n - 1, range[v].first, h[v], ans);
  }
};
vector<int> a, ch[1100000 / 3];
int tm;
void dfs(int v, int p) {
  a.push_back(v);
  h[v] = h[p] + 1;
  range[v].first = tm++;
  for (vector<int>::iterator it = ch[v].begin(), itE = ch[v].end(); it != itE;
       ++it)
    dfs(*it, v);
  range[v].second = tm - 1;
}
int main() {
  h[0] = -1;
  tm = 0;
  int n, i, j, k;
  scanf("%d", &n);
  for (i = 2; i <= n; ++i) {
    scanf("%d", &j);
    ch[j].push_back(i);
  }
  dfs(1, 0);
  SegmentTree st(a);
  int q, v, x, t;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d %d %d", &v, &x, &k);
      st.update(v, x, k);
    } else {
      scanf("%d", &v);
      printf("%lld\n", (st.query(v) + 1000000007ll) % 1000000007ll);
    }
  }
  return 0;
}

```