#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int inf = 50000000;
const int maxn = 300010;
vector<int> adj[maxn], dfstree;
int depth[maxn], strt[maxn], en[maxn];
int st[2][maxn * 5], lazy[2][maxn * 5];
void dfs(int cur, int p, int d) {
  depth[cur] = d;
  strt[cur] = dfstree.size();
  dfstree.push_back(cur);
  for (int it : adj[cur]) {
    if (it != p) dfs(it, cur, d + 1);
  }
  en[cur] = dfstree.size() - 1;
}
int add(int a, int b) {
  int ret = (a + b) % mod;
  return ret;
}
int mult(int a, int b) {
  long long ret = a;
  ret = (ret * b) % mod;
  a = ret;
  return a;
}
void propagate_down(int t, int second, int e, int node) {
  if (lazy[t][node]) {
    st[t][node] = (st[t][node] + lazy[t][node]) % mod;
    if (second != e) {
      int lc = 2 * node + 1;
      int rc = lc + 1;
      lazy[t][lc] = (lazy[t][lc] + lazy[t][node]) % mod;
      lazy[t][rc] = (lazy[t][rc] + lazy[t][node]) % mod;
    }
    lazy[t][node] = 0;
  }
  return;
}
void update(int t, int second, int e, int qs, int qe, int node, int val) {
  propagate_down(t, second, e, node);
  if (second > e || second > qe || qs > e) return;
  int mid, lc, rc;
  mid = (second + e) >> 1;
  lc = 2 * node + 1;
  rc = lc + 1;
  if (second >= qs && qe >= e) {
    st[t][node] = (st[t][node] + val) % mod;
    if (second != e) {
      lazy[t][lc] = (lazy[t][lc] + val) % mod;
      lazy[t][rc] = (lazy[t][rc] + val) % mod;
    }
    return;
  }
  update(t, second, mid, qs, qe, lc, val);
  update(t, mid + 1, e, qs, qe, rc, val);
  st[t][node] = (st[t][lc] + st[t][rc]) % mod;
}
int query(int t, int second, int e, int qs, int qe, int node) {
  propagate_down(t, second, e, node);
  if (second > e || second > qe || qs > e) return 0;
  if (second >= qs && qe >= e) return st[t][node];
  int mid, lc, rc;
  mid = (second + e) >> 1;
  lc = 2 * node + 1;
  rc = lc + 1;
  return add(query(t, second, mid, qs, qe, lc),
             query(t, mid + 1, e, qs, qe, rc));
}
int main() {
  int q, t, u, v, x, k, ans, n, i, j;
  int temp1, temp2;
  scanf("%d", &n);
  for (i = 2; i <= n; i++) {
    scanf("%d", &u);
    adj[u].push_back(i);
    adj[i].push_back(u);
  }
  dfs(1, 0, 0);
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &t, &v);
    if (t == 1) {
      scanf("%d%d", &x, &k);
      update(0, 0, n - 1, strt[v], en[v], 0, k);
      update(1, 0, n - 1, strt[v], en[v], 0, add(mult(depth[v], k), x));
    } else {
      temp1 = query(0, 0, n - 1, strt[v], strt[v], 0);
      temp2 = query(1, 0, n - 1, strt[v], strt[v], 0);
      ans = (temp2 - mult(temp1, depth[v]) + mod) % mod;
      printf("%d\n", ans);
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 mod = 1000000007;
const int inf = 50000000;
const int maxn = 300010;
vector<int> adj[maxn], dfstree;
int depth[maxn], strt[maxn], en[maxn];
int st[2][maxn * 5], lazy[2][maxn * 5];
void dfs(int cur, int p, int d) {
  depth[cur] = d;
  strt[cur] = dfstree.size();
  dfstree.push_back(cur);
  for (int it : adj[cur]) {
    if (it != p) dfs(it, cur, d + 1);
  }
  en[cur] = dfstree.size() - 1;
}
int add(int a, int b) {
  int ret = (a + b) % mod;
  return ret;
}
int mult(int a, int b) {
  long long ret = a;
  ret = (ret * b) % mod;
  a = ret;
  return a;
}
void propagate_down(int t, int second, int e, int node) {
  if (lazy[t][node]) {
    st[t][node] = (st[t][node] + lazy[t][node]) % mod;
    if (second != e) {
      int lc = 2 * node + 1;
      int rc = lc + 1;
      lazy[t][lc] = (lazy[t][lc] + lazy[t][node]) % mod;
      lazy[t][rc] = (lazy[t][rc] + lazy[t][node]) % mod;
    }
    lazy[t][node] = 0;
  }
  return;
}
void update(int t, int second, int e, int qs, int qe, int node, int val) {
  propagate_down(t, second, e, node);
  if (second > e || second > qe || qs > e) return;
  int mid, lc, rc;
  mid = (second + e) >> 1;
  lc = 2 * node + 1;
  rc = lc + 1;
  if (second >= qs && qe >= e) {
    st[t][node] = (st[t][node] + val) % mod;
    if (second != e) {
      lazy[t][lc] = (lazy[t][lc] + val) % mod;
      lazy[t][rc] = (lazy[t][rc] + val) % mod;
    }
    return;
  }
  update(t, second, mid, qs, qe, lc, val);
  update(t, mid + 1, e, qs, qe, rc, val);
  st[t][node] = (st[t][lc] + st[t][rc]) % mod;
}
int query(int t, int second, int e, int qs, int qe, int node) {
  propagate_down(t, second, e, node);
  if (second > e || second > qe || qs > e) return 0;
  if (second >= qs && qe >= e) return st[t][node];
  int mid, lc, rc;
  mid = (second + e) >> 1;
  lc = 2 * node + 1;
  rc = lc + 1;
  return add(query(t, second, mid, qs, qe, lc),
             query(t, mid + 1, e, qs, qe, rc));
}
int main() {
  int q, t, u, v, x, k, ans, n, i, j;
  int temp1, temp2;
  scanf("%d", &n);
  for (i = 2; i <= n; i++) {
    scanf("%d", &u);
    adj[u].push_back(i);
    adj[i].push_back(u);
  }
  dfs(1, 0, 0);
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &t, &v);
    if (t == 1) {
      scanf("%d%d", &x, &k);
      update(0, 0, n - 1, strt[v], en[v], 0, k);
      update(1, 0, n - 1, strt[v], en[v], 0, add(mult(depth[v], k), x));
    } else {
      temp1 = query(0, 0, n - 1, strt[v], strt[v], 0);
      temp2 = query(1, 0, n - 1, strt[v], strt[v], 0);
      ans = (temp2 - mult(temp1, depth[v]) + mod) % mod;
      printf("%d\n", ans);
    }
  }
  return 0;
}

```