#include <bits/stdc++.h>
using namespace std;
const long long M = 1000000007;
struct edge {
  int v, next;
};
edge e[300010 * 2];
int t1[300010 * 4], t2[300010 * 4], h[300010], c[300010], rb[300010], l[300010],
    n, color;
void add(int* t, int v, int tl, int tr, int l, int r, int k) {
  if (l > r) return;
  if (l == tl && r == tr) {
    t[v] = (t[v] + k) % M;
  } else {
    int tm = (tl + tr) / 2;
    add(t, v * 2, tl, tm, l, min(r, tm), k);
    add(t, v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, k);
  }
}
int get(int* t, int v, int tl, int tr, int pos) {
  if (tl == tr) return t[v];
  int tm = (tl + tr) / 2;
  if (pos <= tm)
    return (t[v] + get(t, v * 2, tl, tm, pos)) % M;
  else
    return (t[v] + get(t, v * 2 + 1, tm + 1, tr, pos)) % M;
}
void dfs(int v, int lvl) {
  l[v] = lvl;
  c[v] = color;
  for (int j = h[v]; j > 0; j = e[j].next) {
    int vto = e[j].v;
    if (c[vto] == 0) {
      color++;
      dfs(vto, lvl + 1);
    }
  }
  rb[v] = color;
}
int main() {
  scanf("%d", &n);
  int m = 0;
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    m++;
    e[m].v = i;
    e[m].next = h[x];
    h[x] = m;
    m++;
    e[m].v = x;
    e[m].next = h[i];
    h[i] = m;
  }
  color = 1;
  dfs(1, 1);
  int nq;
  scanf("%d", &nq);
  for (int q = 1; q <= nq; q++) {
    int t, v, x, k;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &v, &x, &k);
      add(t1, 1, 1, n, c[v], rb[v],
          ((long long)x + (long long)l[v] * (long long)k) % M);
      add(t2, 1, 1, n, c[v], rb[v], k);
    } else {
      scanf("%d", &v);
      int k1 = get(t1, 1, 1, n, c[v]);
      int k2 = get(t2, 1, 1, n, c[v]);
      long long ans = ((long long)k1 - (long long)k2 * (long long)l[v]) % M;
      if (ans < 0) ans += M;
      printf("%I64d\n", ans);
    }
  }
  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 long long M = 1000000007;
struct edge {
  int v, next;
};
edge e[300010 * 2];
int t1[300010 * 4], t2[300010 * 4], h[300010], c[300010], rb[300010], l[300010],
    n, color;
void add(int* t, int v, int tl, int tr, int l, int r, int k) {
  if (l > r) return;
  if (l == tl && r == tr) {
    t[v] = (t[v] + k) % M;
  } else {
    int tm = (tl + tr) / 2;
    add(t, v * 2, tl, tm, l, min(r, tm), k);
    add(t, v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, k);
  }
}
int get(int* t, int v, int tl, int tr, int pos) {
  if (tl == tr) return t[v];
  int tm = (tl + tr) / 2;
  if (pos <= tm)
    return (t[v] + get(t, v * 2, tl, tm, pos)) % M;
  else
    return (t[v] + get(t, v * 2 + 1, tm + 1, tr, pos)) % M;
}
void dfs(int v, int lvl) {
  l[v] = lvl;
  c[v] = color;
  for (int j = h[v]; j > 0; j = e[j].next) {
    int vto = e[j].v;
    if (c[vto] == 0) {
      color++;
      dfs(vto, lvl + 1);
    }
  }
  rb[v] = color;
}
int main() {
  scanf("%d", &n);
  int m = 0;
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    m++;
    e[m].v = i;
    e[m].next = h[x];
    h[x] = m;
    m++;
    e[m].v = x;
    e[m].next = h[i];
    h[i] = m;
  }
  color = 1;
  dfs(1, 1);
  int nq;
  scanf("%d", &nq);
  for (int q = 1; q <= nq; q++) {
    int t, v, x, k;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &v, &x, &k);
      add(t1, 1, 1, n, c[v], rb[v],
          ((long long)x + (long long)l[v] * (long long)k) % M);
      add(t2, 1, 1, n, c[v], rb[v], k);
    } else {
      scanf("%d", &v);
      int k1 = get(t1, 1, 1, n, c[v]);
      int k2 = get(t2, 1, 1, n, c[v]);
      long long ans = ((long long)k1 - (long long)k2 * (long long)l[v]) % M;
      if (ans < 0) ans += M;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}

```