#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
const int MOD = 1000000007;
int n, m;
int it[2][N * 4], lazy[2][N * 4], T[N], pos[N], H[N];
vector<int> adj[N];
void dfs(int u, int pa) {
  int L = 1000000000;
  H[u] = H[pa] + 1;
  for (int i = 0; i < (int)adj[u].size(); ++i) {
    int v = adj[u][i];
    if (v == pa) continue;
    dfs(v, u);
    L = min(L, T[v]);
  }
  ++m;
  pos[u] = m;
  T[u] = min(m, L);
}
void lazy_update(int type, int p, int l, int r) {
  if (lazy[type][p] != 0) {
    if (l == r)
      it[type][p] = (it[type][p] + lazy[type][p]) % MOD;
    else {
      lazy[type][p * 2] = (lazy[type][p * 2] + lazy[type][p]) % MOD;
      lazy[type][p * 2 + 1] = (lazy[type][p * 2 + 1] + lazy[type][p]) % MOD;
    }
    lazy[type][p] = 0;
  }
}
void update(int type, int p, int l, int r, int u, int v, int k) {
  lazy_update(type, p, l, r);
  if (l > v || r < u) return;
  if (l >= u && r <= v) {
    lazy[type][p] = (lazy[type][p] + k) % MOD;
    lazy_update(type, p, l, r);
    return;
  }
  int m = (l + r) >> 1;
  update(type, p * 2, l, m, u, v, k);
  update(type, p * 2 + 1, m + 1, r, u, v, k);
}
int retrieve(int type, int p, int l, int r, int u) {
  lazy_update(type, p, l, r);
  if (l > u || r < u) return 0;
  if (l == r) return it[type][p];
  int m = (l + r) >> 1;
  int q1 = retrieve(type, p * 2, l, m, u),
      q2 = retrieve(type, p * 2 + 1, m + 1, r, u);
  return (q1 + q2) % MOD;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  dfs(1, 0);
  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);
      update(0, 1, 1, n, T[v], pos[v], (x + (long long)k * H[v] % MOD) % MOD);
      update(1, 1, 1, n, T[v], pos[v], k);
    } else {
      int v;
      scanf("%d", &v);
      int res = (retrieve(0, 1, 1, n, pos[v]) -
                 (long long)retrieve(1, 1, 1, n, pos[v]) * H[v] % MOD) %
                MOD;
      if (res < 0) res += MOD;
      printf("%d\n", res);
    }
  }
  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;
const int N = 300005;
const int MOD = 1000000007;
int n, m;
int it[2][N * 4], lazy[2][N * 4], T[N], pos[N], H[N];
vector<int> adj[N];
void dfs(int u, int pa) {
  int L = 1000000000;
  H[u] = H[pa] + 1;
  for (int i = 0; i < (int)adj[u].size(); ++i) {
    int v = adj[u][i];
    if (v == pa) continue;
    dfs(v, u);
    L = min(L, T[v]);
  }
  ++m;
  pos[u] = m;
  T[u] = min(m, L);
}
void lazy_update(int type, int p, int l, int r) {
  if (lazy[type][p] != 0) {
    if (l == r)
      it[type][p] = (it[type][p] + lazy[type][p]) % MOD;
    else {
      lazy[type][p * 2] = (lazy[type][p * 2] + lazy[type][p]) % MOD;
      lazy[type][p * 2 + 1] = (lazy[type][p * 2 + 1] + lazy[type][p]) % MOD;
    }
    lazy[type][p] = 0;
  }
}
void update(int type, int p, int l, int r, int u, int v, int k) {
  lazy_update(type, p, l, r);
  if (l > v || r < u) return;
  if (l >= u && r <= v) {
    lazy[type][p] = (lazy[type][p] + k) % MOD;
    lazy_update(type, p, l, r);
    return;
  }
  int m = (l + r) >> 1;
  update(type, p * 2, l, m, u, v, k);
  update(type, p * 2 + 1, m + 1, r, u, v, k);
}
int retrieve(int type, int p, int l, int r, int u) {
  lazy_update(type, p, l, r);
  if (l > u || r < u) return 0;
  if (l == r) return it[type][p];
  int m = (l + r) >> 1;
  int q1 = retrieve(type, p * 2, l, m, u),
      q2 = retrieve(type, p * 2 + 1, m + 1, r, u);
  return (q1 + q2) % MOD;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  dfs(1, 0);
  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);
      update(0, 1, 1, n, T[v], pos[v], (x + (long long)k * H[v] % MOD) % MOD);
      update(1, 1, 1, n, T[v], pos[v], k);
    } else {
      int v;
      scanf("%d", &v);
      int res = (retrieve(0, 1, 1, n, pos[v]) -
                 (long long)retrieve(1, 1, 1, n, pos[v]) * H[v] % MOD) %
                MOD;
      if (res < 0) res += MOD;
      printf("%d\n", res);
    }
  }
  return 0;
}

```