#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;
struct node {
  long long val, to;
} t[2400001], t1[2400001];
vector<int> g[300001];
int tin[300001];
int timer;
int lvl[300001];
int tout[300001];
inline void dfs(int v) {
  tin[v] = ++timer;
  for (int i = 0; i < g[v].size(); i++) {
    if (!tin[g[v][i]]) {
      lvl[g[v][i]] = lvl[v] + 1;
      dfs(g[v][i]);
    }
  }
  tout[v] = timer;
}
int add(int x, int y) {
  long long res = x + y;
  if (res < 0) res += mod;
  if (res >= mod) res -= mod;
  return res;
}
int mult(int x, int y) {
  long long res = 1ll * x * y;
  while (res < 0) res += mod;
  if (res >= mod) res = res % mod;
  return res;
}
inline void push1(int v, int tl, int tr) {
  if (t1[v].to == 0) return;
  t1[v].val = add(t1[v].val, mult(t1[v].to, (tr - tl + 1) * 1ll));
  if (tl != tr) {
    t1[v + v].to = add(t1[v + v].to, t1[v].to);
    t1[v + v + 1].to = add(t1[v + v + 1].to, t1[v].to);
  }
  t1[v].to = 0;
}
inline void upd1(int l, int r, int val, int v = 1, int tl = 1, int tr = timer) {
  push1(v, tl, tr);
  if (tr < l || tl > r) return;
  if (tl >= l && tr <= r) {
    t1[v].to += val;
    push1(v, tl, tr);
    return;
  }
  int tm = (tl + tr) >> 1;
  upd1(l, r, val, v + v, tl, tm);
  upd1(l, r, val, v + v + 1, tm + 1, tr);
  t1[v].val = add(t1[v + v].val, t1[v + v + 1].val);
}
int get1(int l, int r, int v = 1, int tl = 1, int tr = timer) {
  push1(v, tl, tr);
  if (tr < l || tl > r) return 0;
  if (tl >= l && tr <= r) return t1[v].val;
  int tm = (tl + tr) >> 1;
  return add(get1(l, r, v + v, tl, tm), get1(l, r, v + v + 1, tm + 1, tr));
}
inline void push(int v, int tl, int tr) {
  if (t[v].to == 0) return;
  t[v].val = add(t[v].val, mult(t[v].to, tr - tl + 1));
  if (tl != tr) {
    t[v + v].to = add(t[v + v].to, t[v].to);
    t[v + v + 1].to = add(t[v + v + 1].to, t[v].to);
  }
  t[v].to = 0;
}
inline void upd(int l, int r, int val, int v = 1, int tl = 1, int tr = timer) {
  push(v, tl, tr);
  if (tr < l || tl > r) return;
  if (tl >= l && tr <= r) {
    t[v].to += val;
    push(v, tl, tr);
    return;
  }
  int tm = (tl + tr) >> 1;
  upd(l, r, val, v + v, tl, tm);
  upd(l, r, val, v + v + 1, tm + 1, tr);
  t[v].val = add(t[v + v].val, t[v + v + 1].val);
}
int get(int l, int r, int v = 1, int tl = 1, int tr = timer) {
  push(v, tl, tr);
  if (tr < l || tl > r) return 0;
  if (tl >= l && tr <= r) return t[v].val;
  int tm = (tl + tr) >> 1;
  return add(get(l, r, v + v, tl, tm), get(l, r, v + v + 1, tm + 1, tr));
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    g[x].push_back(i);
    g[i].push_back(x);
  }
  dfs(1);
  int q;
  cin >> q;
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int v, k, x;
      scanf("%d%d%d", &v, &x, &k);
      int jkl = add(x, mult(lvl[v], k));
      upd(tin[v], tout[v], jkl);
      upd1(tin[v], tout[v], k);
    } else {
      int v;
      scanf("%d", &v);
      int ans = add(get(tin[v], tin[v]), -mult(get1(tin[v], tin[v]), lvl[v]));
      printf("%d\n", ans);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int mod = 1e9 + 7;
struct node {
  long long val, to;
} t[2400001], t1[2400001];
vector<int> g[300001];
int tin[300001];
int timer;
int lvl[300001];
int tout[300001];
inline void dfs(int v) {
  tin[v] = ++timer;
  for (int i = 0; i < g[v].size(); i++) {
    if (!tin[g[v][i]]) {
      lvl[g[v][i]] = lvl[v] + 1;
      dfs(g[v][i]);
    }
  }
  tout[v] = timer;
}
int add(int x, int y) {
  long long res = x + y;
  if (res < 0) res += mod;
  if (res >= mod) res -= mod;
  return res;
}
int mult(int x, int y) {
  long long res = 1ll * x * y;
  while (res < 0) res += mod;
  if (res >= mod) res = res % mod;
  return res;
}
inline void push1(int v, int tl, int tr) {
  if (t1[v].to == 0) return;
  t1[v].val = add(t1[v].val, mult(t1[v].to, (tr - tl + 1) * 1ll));
  if (tl != tr) {
    t1[v + v].to = add(t1[v + v].to, t1[v].to);
    t1[v + v + 1].to = add(t1[v + v + 1].to, t1[v].to);
  }
  t1[v].to = 0;
}
inline void upd1(int l, int r, int val, int v = 1, int tl = 1, int tr = timer) {
  push1(v, tl, tr);
  if (tr < l || tl > r) return;
  if (tl >= l && tr <= r) {
    t1[v].to += val;
    push1(v, tl, tr);
    return;
  }
  int tm = (tl + tr) >> 1;
  upd1(l, r, val, v + v, tl, tm);
  upd1(l, r, val, v + v + 1, tm + 1, tr);
  t1[v].val = add(t1[v + v].val, t1[v + v + 1].val);
}
int get1(int l, int r, int v = 1, int tl = 1, int tr = timer) {
  push1(v, tl, tr);
  if (tr < l || tl > r) return 0;
  if (tl >= l && tr <= r) return t1[v].val;
  int tm = (tl + tr) >> 1;
  return add(get1(l, r, v + v, tl, tm), get1(l, r, v + v + 1, tm + 1, tr));
}
inline void push(int v, int tl, int tr) {
  if (t[v].to == 0) return;
  t[v].val = add(t[v].val, mult(t[v].to, tr - tl + 1));
  if (tl != tr) {
    t[v + v].to = add(t[v + v].to, t[v].to);
    t[v + v + 1].to = add(t[v + v + 1].to, t[v].to);
  }
  t[v].to = 0;
}
inline void upd(int l, int r, int val, int v = 1, int tl = 1, int tr = timer) {
  push(v, tl, tr);
  if (tr < l || tl > r) return;
  if (tl >= l && tr <= r) {
    t[v].to += val;
    push(v, tl, tr);
    return;
  }
  int tm = (tl + tr) >> 1;
  upd(l, r, val, v + v, tl, tm);
  upd(l, r, val, v + v + 1, tm + 1, tr);
  t[v].val = add(t[v + v].val, t[v + v + 1].val);
}
int get(int l, int r, int v = 1, int tl = 1, int tr = timer) {
  push(v, tl, tr);
  if (tr < l || tl > r) return 0;
  if (tl >= l && tr <= r) return t[v].val;
  int tm = (tl + tr) >> 1;
  return add(get(l, r, v + v, tl, tm), get(l, r, v + v + 1, tm + 1, tr));
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    g[x].push_back(i);
    g[i].push_back(x);
  }
  dfs(1);
  int q;
  cin >> q;
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int v, k, x;
      scanf("%d%d%d", &v, &x, &k);
      int jkl = add(x, mult(lvl[v], k));
      upd(tin[v], tout[v], jkl);
      upd1(tin[v], tout[v], k);
    } else {
      int v;
      scanf("%d", &v);
      int ans = add(get(tin[v], tin[v]), -mult(get1(tin[v], tin[v]), lvl[v]));
      printf("%d\n", ans);
    }
  }
  return 0;
}

```