#include <bits/stdc++.h>
using namespace std;
const int maxn = 3 * 1e5 + 3;
const long long mod = 1e9 + 7;
int n, m, in[maxn], out[maxn], cnt = 0, avail[maxn];
long long lev[maxn], hkx[4 * maxn], sk[4 * maxn];
vector<int> g[maxn];
void DFS(int u) {
  avail[u] = 1;
  in[u] = ++cnt;
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (avail[v] == 0) {
      lev[v] = lev[u] + 1;
      DFS(v);
    }
  }
  out[u] = cnt;
}
void Update(int node, int start, int endd, int v, long long x, long long k) {
  if (in[v] > endd || out[v] < start || start > endd) {
    return;
  }
  if (in[v] <= start && endd <= out[v]) {
    hkx[node] = (hkx[node] + x + k * lev[v]) % mod;
    sk[node] = (sk[node] + k) % mod;
    return;
  }
  int mid = (start + endd) / 2;
  Update(2 * node, start, mid, v, x, k);
  Update(2 * node + 1, mid + 1, endd, v, x, k);
}
long long Ans(int node, int start, int endd, int v) {
  if (in[v] > endd || in[v] < start) {
    return 0;
  }
  long long res = hkx[node] % mod;
  res = (res + mod * mod - lev[v] * sk[node]) % mod;
  if (start == endd) {
    return (res % mod);
  }
  int mid = (start + endd) / 2;
  long long l = Ans(2 * node, start, mid, v) % mod;
  long long r = Ans(2 * node + 1, mid + 1, endd, v) % mod;
  return ((res + l + r) % mod);
}
void BuildGraph() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int p;
    scanf("%d", &p);
    g[p].push_back(i), g[i].push_back(p);
  }
}
void Query() {
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int v;
      long long x, k;
      scanf("%d%lld%lld", &v, &x, &k);
      Update(1, 1, n, v, x, k);
    }
    if (type == 2) {
      int v;
      scanf("%d", &v);
      printf("%lld\n", (Ans(1, 1, n, v) % mod));
    }
  }
}
int main() {
  BuildGraph();
  lev[1] = 1;
  DFS(1);
  Query();
}

### 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;
const int maxn = 3 * 1e5 + 3;
const long long mod = 1e9 + 7;
int n, m, in[maxn], out[maxn], cnt = 0, avail[maxn];
long long lev[maxn], hkx[4 * maxn], sk[4 * maxn];
vector<int> g[maxn];
void DFS(int u) {
  avail[u] = 1;
  in[u] = ++cnt;
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (avail[v] == 0) {
      lev[v] = lev[u] + 1;
      DFS(v);
    }
  }
  out[u] = cnt;
}
void Update(int node, int start, int endd, int v, long long x, long long k) {
  if (in[v] > endd || out[v] < start || start > endd) {
    return;
  }
  if (in[v] <= start && endd <= out[v]) {
    hkx[node] = (hkx[node] + x + k * lev[v]) % mod;
    sk[node] = (sk[node] + k) % mod;
    return;
  }
  int mid = (start + endd) / 2;
  Update(2 * node, start, mid, v, x, k);
  Update(2 * node + 1, mid + 1, endd, v, x, k);
}
long long Ans(int node, int start, int endd, int v) {
  if (in[v] > endd || in[v] < start) {
    return 0;
  }
  long long res = hkx[node] % mod;
  res = (res + mod * mod - lev[v] * sk[node]) % mod;
  if (start == endd) {
    return (res % mod);
  }
  int mid = (start + endd) / 2;
  long long l = Ans(2 * node, start, mid, v) % mod;
  long long r = Ans(2 * node + 1, mid + 1, endd, v) % mod;
  return ((res + l + r) % mod);
}
void BuildGraph() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int p;
    scanf("%d", &p);
    g[p].push_back(i), g[i].push_back(p);
  }
}
void Query() {
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int v;
      long long x, k;
      scanf("%d%lld%lld", &v, &x, &k);
      Update(1, 1, n, v, x, k);
    }
    if (type == 2) {
      int v;
      scanf("%d", &v);
      printf("%lld\n", (Ans(1, 1, n, v) % mod));
    }
  }
}
int main() {
  BuildGraph();
  lev[1] = 1;
  DFS(1);
  Query();
}

```