#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
const double eps = 1e-15;
const double pi = acos(-1.0);
const int MAXN = (int)3e5 + 7;
const int INF = (long long)2e9;
const long long LINF = (long long)8e18;
const int MOD = (long long)1e9 + 7;
int solve();
int main() {
  solve();
  return 0;
}
int n, q, nn;
vector<int> g[MAXN];
vector<int> order;
long long leftt[MAXN], rightt[MAXN], h[MAXN], t1[MAXN << 2], t2[MAXN << 2];
long long start[MAXN];
bool used[MAXN];
void dfs(int v, int deep) {
  h[v] = deep;
  leftt[v] = order.size();
  order.push_back(v);
  for (int to : g[v]) {
    dfs(to, deep + 1);
  }
  rightt[v] = order.size();
  order.push_back(v);
}
void update(long long *t, int pos, long long inc) {
  for (int i = pos; i < nn; i |= i + 1) {
    t[i] += inc;
    t[i] = (t[i] + MOD) % MOD;
  }
}
void add(long long *t, int l, int r, long long inc) {
  update(t, l, inc);
  update(t, r + 1, -inc);
}
long long get(long long *t, int pos) {
  long long ans = 0;
  for (int i = pos; i >= 0; i = i = (i & (i + 1)) - 1) {
    ans += t[i];
    ans %= MOD;
  }
  return ans;
}
int ask(int v) {
  long long s1 = get(t1, leftt[v]);
  long long s2 = get(t2, leftt[v]);
  return (int)((2 * MOD + s1 - (s2 * h[v]) % MOD) % MOD);
}
int solve() {
  scanf("%d ", &n);
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    g[x].push_back(i);
  }
  dfs(1, 0);
  nn = order.size() - 1;
  scanf("%d", &q);
  for (int test = 0; test < (int)q; test++) {
    int v, type;
    scanf("%d%d", &type, &v);
    if (type == 1) {
      int k, x;
      scanf("%d%d\n", &x, &k);
      long long inc = ((long long)x + (long long)h[v] * (long long)k) % MOD;
      add(t1, leftt[v], rightt[v], inc);
      add(t2, leftt[v], rightt[v], k);
    } else {
      printf("%d\n", ask(v));
    }
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
const double eps = 1e-15;
const double pi = acos(-1.0);
const int MAXN = (int)3e5 + 7;
const int INF = (long long)2e9;
const long long LINF = (long long)8e18;
const int MOD = (long long)1e9 + 7;
int solve();
int main() {
  solve();
  return 0;
}
int n, q, nn;
vector<int> g[MAXN];
vector<int> order;
long long leftt[MAXN], rightt[MAXN], h[MAXN], t1[MAXN << 2], t2[MAXN << 2];
long long start[MAXN];
bool used[MAXN];
void dfs(int v, int deep) {
  h[v] = deep;
  leftt[v] = order.size();
  order.push_back(v);
  for (int to : g[v]) {
    dfs(to, deep + 1);
  }
  rightt[v] = order.size();
  order.push_back(v);
}
void update(long long *t, int pos, long long inc) {
  for (int i = pos; i < nn; i |= i + 1) {
    t[i] += inc;
    t[i] = (t[i] + MOD) % MOD;
  }
}
void add(long long *t, int l, int r, long long inc) {
  update(t, l, inc);
  update(t, r + 1, -inc);
}
long long get(long long *t, int pos) {
  long long ans = 0;
  for (int i = pos; i >= 0; i = i = (i & (i + 1)) - 1) {
    ans += t[i];
    ans %= MOD;
  }
  return ans;
}
int ask(int v) {
  long long s1 = get(t1, leftt[v]);
  long long s2 = get(t2, leftt[v]);
  return (int)((2 * MOD + s1 - (s2 * h[v]) % MOD) % MOD);
}
int solve() {
  scanf("%d ", &n);
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    g[x].push_back(i);
  }
  dfs(1, 0);
  nn = order.size() - 1;
  scanf("%d", &q);
  for (int test = 0; test < (int)q; test++) {
    int v, type;
    scanf("%d%d", &type, &v);
    if (type == 1) {
      int k, x;
      scanf("%d%d\n", &x, &k);
      long long inc = ((long long)x + (long long)h[v] * (long long)k) % MOD;
      add(t1, leftt[v], rightt[v], inc);
      add(t2, leftt[v], rightt[v], k);
    } else {
      printf("%d\n", ask(v));
    }
  }
  return 0;
}

```