#include <bits/stdc++.h>
using namespace std;
using ll = int_fast64_t;
const int M = (int)1e9 + 7;
void dfs(int v, vector<vector<int> > &g, vector<int> &h, vector<int> &order,
         vector<int> &cnt) {
  order.push_back(v);
  cnt[v] = 0;
  for (auto to : g[v]) {
    h[to] = h[v] + 1;
    dfs(to, g, h, order, cnt);
    cnt[v] += cnt[to] + 1;
  }
}
struct tree {
  vector<int> add;
  tree(int n) : add(4 * n) {}
  void mod(int v, int l, int r, int val, int tl, int tr) {
    if (l < tl) {
      l = tl;
    }
    if (r > tr) {
      r = tr;
    }
    if (l > r) {
      return;
    }
    if (l == tl && r == tr) {
      add[v] += val;
      if (add[v] > M) {
        add[v] -= M;
      }
      return;
    }
    add[v * 2] += add[v];
    if (add[v * 2] >= M) {
      add[v * 2] -= M;
    }
    add[v * 2 + 1] += add[v];
    if (add[v * 2 + 1] >= M) {
      add[v * 2 + 1] -= M;
    }
    add[v] = 0;
    mod(v * 2, l, r, val, tl, (tl + tr) / 2);
    mod(v * 2 + 1, l, r, val, (tl + tr) / 2 + 1, tr);
  }
  int get(int v, int x, int tl, int tr) {
    if (tl == tr) {
      return add[v];
    }
    add[v * 2] += add[v];
    if (add[v * 2] >= M) {
      add[v * 2] -= M;
    }
    add[v * 2 + 1] += add[v];
    if (add[v * 2 + 1] >= M) {
      add[v * 2 + 1] -= M;
    }
    add[v] = 0;
    if (x <= (tl + tr) / 2) {
      return get(v * 2, x, tl, (tl + tr) / 2);
    } else {
      return get(v * 2 + 1, x, (tl + tr) / 2 + 1, tr);
    }
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> p(n);
  vector<vector<int> > g(n);
  for (int i = 1; i < n; ++i) {
    cin >> p[i];
    --p[i];
    g[p[i]].push_back(i);
  }
  vector<int> order, h(n), cnt(n);
  dfs(0, g, h, order, cnt);
  vector<int> pos(n);
  for (int i = 0; i < n; ++i) {
    pos[order[i]] = i;
  }
  tree x(n), k(n);
  int q;
  cin >> q;
  while (q-- > 0) {
    int type;
    cin >> type;
    if (--type) {
      int v;
      cin >> v;
      --v;
      int ans =
          (x.get(1, pos[v], 0, n - 1) - (ll)h[v] * k.get(1, pos[v], 0, n - 1)) %
          M;
      if (ans < 0) {
        ans += M;
      }
      cout << ans << '\n';
    } else {
      int v, X, K;
      cin >> v >> X >> K;
      --v;
      x.mod(1, pos[v], pos[v] + cnt[v], ((X + (ll)h[v] * K) % M + M) % M, 0,
            n - 1);
      k.mod(1, pos[v], pos[v] + cnt[v], K, 0, n - 1);
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using ll = int_fast64_t;
const int M = (int)1e9 + 7;
void dfs(int v, vector<vector<int> > &g, vector<int> &h, vector<int> &order,
         vector<int> &cnt) {
  order.push_back(v);
  cnt[v] = 0;
  for (auto to : g[v]) {
    h[to] = h[v] + 1;
    dfs(to, g, h, order, cnt);
    cnt[v] += cnt[to] + 1;
  }
}
struct tree {
  vector<int> add;
  tree(int n) : add(4 * n) {}
  void mod(int v, int l, int r, int val, int tl, int tr) {
    if (l < tl) {
      l = tl;
    }
    if (r > tr) {
      r = tr;
    }
    if (l > r) {
      return;
    }
    if (l == tl && r == tr) {
      add[v] += val;
      if (add[v] > M) {
        add[v] -= M;
      }
      return;
    }
    add[v * 2] += add[v];
    if (add[v * 2] >= M) {
      add[v * 2] -= M;
    }
    add[v * 2 + 1] += add[v];
    if (add[v * 2 + 1] >= M) {
      add[v * 2 + 1] -= M;
    }
    add[v] = 0;
    mod(v * 2, l, r, val, tl, (tl + tr) / 2);
    mod(v * 2 + 1, l, r, val, (tl + tr) / 2 + 1, tr);
  }
  int get(int v, int x, int tl, int tr) {
    if (tl == tr) {
      return add[v];
    }
    add[v * 2] += add[v];
    if (add[v * 2] >= M) {
      add[v * 2] -= M;
    }
    add[v * 2 + 1] += add[v];
    if (add[v * 2 + 1] >= M) {
      add[v * 2 + 1] -= M;
    }
    add[v] = 0;
    if (x <= (tl + tr) / 2) {
      return get(v * 2, x, tl, (tl + tr) / 2);
    } else {
      return get(v * 2 + 1, x, (tl + tr) / 2 + 1, tr);
    }
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> p(n);
  vector<vector<int> > g(n);
  for (int i = 1; i < n; ++i) {
    cin >> p[i];
    --p[i];
    g[p[i]].push_back(i);
  }
  vector<int> order, h(n), cnt(n);
  dfs(0, g, h, order, cnt);
  vector<int> pos(n);
  for (int i = 0; i < n; ++i) {
    pos[order[i]] = i;
  }
  tree x(n), k(n);
  int q;
  cin >> q;
  while (q-- > 0) {
    int type;
    cin >> type;
    if (--type) {
      int v;
      cin >> v;
      --v;
      int ans =
          (x.get(1, pos[v], 0, n - 1) - (ll)h[v] * k.get(1, pos[v], 0, n - 1)) %
          M;
      if (ans < 0) {
        ans += M;
      }
      cout << ans << '\n';
    } else {
      int v, X, K;
      cin >> v >> X >> K;
      --v;
      x.mod(1, pos[v], pos[v] + cnt[v], ((X + (ll)h[v] * K) % M + M) % M, 0,
            n - 1);
      k.mod(1, pos[v], pos[v] + cnt[v], K, 0, n - 1);
    }
  }
  return 0;
}

```