#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 6.66;
const long long MOD = 1e9 + 7;
long long sum[MAXN * 4];
long long lazy_k[MAXN * 4];
long long lazy_x[MAXN * 4];
int st[MAXN], ft[MAXN];
int h[MAXN], tl[MAXN];
long long hpar[MAXN];
int now = 0;
vector<int> tree[MAXN];
void dfs(int v, int hei = 0) {
  st[v] = ++now;
  tl[now] = v;
  h[now] = hei;
  for (auto u : tree[v]) dfs(u, hei + 1);
  ft[v] = now;
}
void shift(int idx, int l, int r) {
  if (lazy_x[idx] == lazy_k[idx] && lazy_k[idx] == 0) return;
  int mid = l + (r - l) / 2;
  lazy_k[idx * 2] = (lazy_k[idx * 2] + lazy_k[idx]) % MOD;
  lazy_x[idx * 2] = (lazy_x[idx * 2] + lazy_x[idx]) % MOD;
  sum[idx * 2] =
      (sum[idx * 2] + (1LL * (mid - l) * lazy_x[idx]) % MOD + MOD +
       ((((1LL * (mid - l) * h[l]) % MOD - hpar[mid - 1] + hpar[l - 1] + MOD) %
         MOD) *
        lazy_k[idx]) %
           MOD) %
      MOD;
  lazy_k[idx * 2 + 1] = (lazy_k[idx * 2 + 1] + lazy_k[idx]) % MOD;
  lazy_x[idx * 2 + 1] =
      (lazy_x[idx * 2 + 1] + lazy_x[idx] +
       (1LL * lazy_k[idx] * (MOD + h[l] - h[mid])) % MOD + MOD) %
      MOD;
  sum[idx * 2 + 1] =
      (sum[idx * 2 + 1] +
       (1LL * (r - mid) *
        (lazy_x[idx] +
         (lazy_k[idx] * 1LL * ((h[l] - h[mid] + MOD) % MOD)) % MOD)) %
           MOD +
       MOD +
       ((((1LL * (r - mid) * h[mid]) % MOD - hpar[r - 1] + hpar[mid - 1] +
          MOD) %
         MOD) *
        lazy_k[idx]) %
           MOD) %
      MOD;
  lazy_x[idx] = lazy_k[idx] = 0;
}
long long intersection(int s1, int s2, int b, int e) {
  if (s1 <= e || b >= s2) return 0;
  return h[max(s1, b)] - h[min(e, s2)];
}
void upd(int idx, int l, int r, int b, int e, long long x, long long k) {
  if (l >= r || l >= e || r <= b) return;
  if (l >= b && r <= e) {
    lazy_k[idx] = (lazy_k[idx] + k) % MOD;
    lazy_x[idx] =
        (lazy_x[idx] + x + (1LL * k * ((h[b] - h[l] + MOD)) % MOD) % MOD) % MOD;
    sum[idx] =
        (sum[idx] +
         (1LL * (r - l) * (x + (k * (h[b] - h[l] + MOD) % MOD) % MOD)) % MOD +
         MOD +
         (1LL * (r - l) * h[l] % MOD - hpar[r - 1] + hpar[l - 1] + MOD) * k %
             MOD) %
        MOD;
    return;
  }
  int mid = l + (r - l) / 2;
  shift(idx, l, r);
  upd(idx * 2, l, mid, b, e, x, k);
  upd(idx * 2 + 1, mid, r, b, e, x, k);
  sum[idx] = (sum[idx * 2] + sum[idx * 2 + 1]) % MOD;
}
long long get(int idx, int l, int r, int b, int e) {
  if (b >= e || l >= r) return 0;
  if (l >= e || r <= b) return 0;
  if (l >= b && r <= e) return sum[idx];
  shift(idx, l, r);
  int mid = l + (r - l) / 2;
  return (get(idx * 2, l, mid, b, e) + get(idx * 2 + 1, mid, r, b, e)) % MOD;
}
long long get(int idx, int l, int r, int x) {
  if (l >= r) return 0;
  if (l > x || r <= x) return 0;
  if (r - l == 1) return sum[idx];
  shift(idx, l, r);
  int mid = l + (r - l) / 2;
  return (get(idx * 2, l, mid, x) + get(idx * 2 + 1, mid, r, x));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    tree[x].push_back(i);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) hpar[i] = (hpar[i - 1] + h[i]) % MOD;
  int Q;
  cin >> Q;
  while (Q--) {
    int X;
    cin >> X;
    if (X == 2) {
      int a;
      cin >> a;
      cout << (get(1, 1, n + 1, 1, st[a] + 1) - get(1, 1, n + 1, 1, st[a]) +
               MOD) %
                  MOD
           << "\n";
    } else {
      int a;
      long long b, c;
      cin >> a >> b >> c;
      upd(1, 1, n + 1, st[a], ft[a] + 1, b, c);
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 int MAXN = 3e5 + 6.66;
const long long MOD = 1e9 + 7;
long long sum[MAXN * 4];
long long lazy_k[MAXN * 4];
long long lazy_x[MAXN * 4];
int st[MAXN], ft[MAXN];
int h[MAXN], tl[MAXN];
long long hpar[MAXN];
int now = 0;
vector<int> tree[MAXN];
void dfs(int v, int hei = 0) {
  st[v] = ++now;
  tl[now] = v;
  h[now] = hei;
  for (auto u : tree[v]) dfs(u, hei + 1);
  ft[v] = now;
}
void shift(int idx, int l, int r) {
  if (lazy_x[idx] == lazy_k[idx] && lazy_k[idx] == 0) return;
  int mid = l + (r - l) / 2;
  lazy_k[idx * 2] = (lazy_k[idx * 2] + lazy_k[idx]) % MOD;
  lazy_x[idx * 2] = (lazy_x[idx * 2] + lazy_x[idx]) % MOD;
  sum[idx * 2] =
      (sum[idx * 2] + (1LL * (mid - l) * lazy_x[idx]) % MOD + MOD +
       ((((1LL * (mid - l) * h[l]) % MOD - hpar[mid - 1] + hpar[l - 1] + MOD) %
         MOD) *
        lazy_k[idx]) %
           MOD) %
      MOD;
  lazy_k[idx * 2 + 1] = (lazy_k[idx * 2 + 1] + lazy_k[idx]) % MOD;
  lazy_x[idx * 2 + 1] =
      (lazy_x[idx * 2 + 1] + lazy_x[idx] +
       (1LL * lazy_k[idx] * (MOD + h[l] - h[mid])) % MOD + MOD) %
      MOD;
  sum[idx * 2 + 1] =
      (sum[idx * 2 + 1] +
       (1LL * (r - mid) *
        (lazy_x[idx] +
         (lazy_k[idx] * 1LL * ((h[l] - h[mid] + MOD) % MOD)) % MOD)) %
           MOD +
       MOD +
       ((((1LL * (r - mid) * h[mid]) % MOD - hpar[r - 1] + hpar[mid - 1] +
          MOD) %
         MOD) *
        lazy_k[idx]) %
           MOD) %
      MOD;
  lazy_x[idx] = lazy_k[idx] = 0;
}
long long intersection(int s1, int s2, int b, int e) {
  if (s1 <= e || b >= s2) return 0;
  return h[max(s1, b)] - h[min(e, s2)];
}
void upd(int idx, int l, int r, int b, int e, long long x, long long k) {
  if (l >= r || l >= e || r <= b) return;
  if (l >= b && r <= e) {
    lazy_k[idx] = (lazy_k[idx] + k) % MOD;
    lazy_x[idx] =
        (lazy_x[idx] + x + (1LL * k * ((h[b] - h[l] + MOD)) % MOD) % MOD) % MOD;
    sum[idx] =
        (sum[idx] +
         (1LL * (r - l) * (x + (k * (h[b] - h[l] + MOD) % MOD) % MOD)) % MOD +
         MOD +
         (1LL * (r - l) * h[l] % MOD - hpar[r - 1] + hpar[l - 1] + MOD) * k %
             MOD) %
        MOD;
    return;
  }
  int mid = l + (r - l) / 2;
  shift(idx, l, r);
  upd(idx * 2, l, mid, b, e, x, k);
  upd(idx * 2 + 1, mid, r, b, e, x, k);
  sum[idx] = (sum[idx * 2] + sum[idx * 2 + 1]) % MOD;
}
long long get(int idx, int l, int r, int b, int e) {
  if (b >= e || l >= r) return 0;
  if (l >= e || r <= b) return 0;
  if (l >= b && r <= e) return sum[idx];
  shift(idx, l, r);
  int mid = l + (r - l) / 2;
  return (get(idx * 2, l, mid, b, e) + get(idx * 2 + 1, mid, r, b, e)) % MOD;
}
long long get(int idx, int l, int r, int x) {
  if (l >= r) return 0;
  if (l > x || r <= x) return 0;
  if (r - l == 1) return sum[idx];
  shift(idx, l, r);
  int mid = l + (r - l) / 2;
  return (get(idx * 2, l, mid, x) + get(idx * 2 + 1, mid, r, x));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    tree[x].push_back(i);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) hpar[i] = (hpar[i - 1] + h[i]) % MOD;
  int Q;
  cin >> Q;
  while (Q--) {
    int X;
    cin >> X;
    if (X == 2) {
      int a;
      cin >> a;
      cout << (get(1, 1, n + 1, 1, st[a] + 1) - get(1, 1, n + 1, 1, st[a]) +
               MOD) %
                  MOD
           << "\n";
    } else {
      int a;
      long long b, c;
      cin >> a >> b >> c;
      upd(1, 1, n + 1, st[a], ft[a] + 1, b, c);
    }
  }
  return 0;
}

```