#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int re = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == -1) return -1;
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    re = re * 10 + c - '0';
    c = getchar();
  }
  return re * fh;
}
inline long long readll() {
  long long re = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == -1) return -1;
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    re = re * 10 + c - '0';
    c = getchar();
  }
  return re * fh;
}
template <class T>
inline void write(T x) {
  if (x == 0) return (void)putchar('0');
  if (x < 0) putchar('-'), x = -x;
  static char buf[30];
  char* p = buf;
  while (x) {
    *(p++) = x % 10 + '0';
    x /= 10;
  }
  while (p > buf) putchar(*(--p));
}
template <>
inline void write<char*>(char* s) {
  while (*s != 0) putchar(*(s++));
}
template <>
inline void write<const char*>(const char* s) {
  while (*s != 0) putchar(*(s++));
}
inline int readstr(char* s) {
  char* begin = s;
  char c = getchar();
  while (c < 'a' || c > 'z') c = getchar();
  while (c >= 'a' && c <= 'z') {
    *(s++) = c;
    c = getchar();
  }
  *s = 0;
  return s - begin;
}
int first[300300], nxte[600600], to[600600], dep[300300], idx[300300],
    siz[300300], tot;
int n, m;
struct BT {
  long long t[300300];
  void add(int l, int r, long long v) {
    v %= 1000000007;
    for (; l <= n; l += l & -l) t[l] = (t[l] + v) % 1000000007;
    for (++r; r <= n; r += r & -r) t[r] = (t[r] - v) % 1000000007;
  }
  long long query(int p) {
    int ret = 0;
    for (; p; p -= p & -p) ret = (ret + t[p]) % 1000000007;
    return ret;
  }
} tx, tk;
void add_e(int a, int b) {
  to[++tot] = b;
  nxte[tot] = first[a];
  first[a] = tot;
}
void dfs(int now, int f) {
  dep[now] = dep[f] + 1;
  siz[now] = 1;
  idx[now] = ++idx[0];
  for (int e = first[now], nxt; e; e = nxte[e]) {
    if ((nxt = to[e]) == f) continue;
    dfs(nxt, now);
    siz[now] += siz[nxt];
  }
}
int main() {
  n = read();
  int x, v, k;
  for (int i = 2; i <= n; i++) {
    x = read();
    add_e(i, x);
    add_e(x, i);
  }
  dfs(1, 0);
  m = read();
  for (int i = 1; i <= m; i++) {
    switch (read()) {
      case 1:
        x = read(), v = read(), k = read();
        tx.add(idx[x], idx[x] + siz[x] - 1, v + (long long)dep[x] * k);
        tk.add(idx[x], idx[x] + siz[x] - 1, k);
        break;
      case 2:
        x = read();
        printf("%d\n", (int)((tx.query(idx[x]) - tk.query(idx[x]) * dep[x]) %
                                 1000000007 +
                             1000000007) %
                           1000000007);
        break;
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
inline int read() {
  int re = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == -1) return -1;
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    re = re * 10 + c - '0';
    c = getchar();
  }
  return re * fh;
}
inline long long readll() {
  long long re = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == -1) return -1;
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    re = re * 10 + c - '0';
    c = getchar();
  }
  return re * fh;
}
template <class T>
inline void write(T x) {
  if (x == 0) return (void)putchar('0');
  if (x < 0) putchar('-'), x = -x;
  static char buf[30];
  char* p = buf;
  while (x) {
    *(p++) = x % 10 + '0';
    x /= 10;
  }
  while (p > buf) putchar(*(--p));
}
template <>
inline void write<char*>(char* s) {
  while (*s != 0) putchar(*(s++));
}
template <>
inline void write<const char*>(const char* s) {
  while (*s != 0) putchar(*(s++));
}
inline int readstr(char* s) {
  char* begin = s;
  char c = getchar();
  while (c < 'a' || c > 'z') c = getchar();
  while (c >= 'a' && c <= 'z') {
    *(s++) = c;
    c = getchar();
  }
  *s = 0;
  return s - begin;
}
int first[300300], nxte[600600], to[600600], dep[300300], idx[300300],
    siz[300300], tot;
int n, m;
struct BT {
  long long t[300300];
  void add(int l, int r, long long v) {
    v %= 1000000007;
    for (; l <= n; l += l & -l) t[l] = (t[l] + v) % 1000000007;
    for (++r; r <= n; r += r & -r) t[r] = (t[r] - v) % 1000000007;
  }
  long long query(int p) {
    int ret = 0;
    for (; p; p -= p & -p) ret = (ret + t[p]) % 1000000007;
    return ret;
  }
} tx, tk;
void add_e(int a, int b) {
  to[++tot] = b;
  nxte[tot] = first[a];
  first[a] = tot;
}
void dfs(int now, int f) {
  dep[now] = dep[f] + 1;
  siz[now] = 1;
  idx[now] = ++idx[0];
  for (int e = first[now], nxt; e; e = nxte[e]) {
    if ((nxt = to[e]) == f) continue;
    dfs(nxt, now);
    siz[now] += siz[nxt];
  }
}
int main() {
  n = read();
  int x, v, k;
  for (int i = 2; i <= n; i++) {
    x = read();
    add_e(i, x);
    add_e(x, i);
  }
  dfs(1, 0);
  m = read();
  for (int i = 1; i <= m; i++) {
    switch (read()) {
      case 1:
        x = read(), v = read(), k = read();
        tx.add(idx[x], idx[x] + siz[x] - 1, v + (long long)dep[x] * k);
        tk.add(idx[x], idx[x] + siz[x] - 1, k);
        break;
      case 2:
        x = read();
        printf("%d\n", (int)((tx.query(idx[x]) - tk.query(idx[x]) * dep[x]) %
                                 1000000007 +
                             1000000007) %
                           1000000007);
        break;
    }
  }
  return 0;
}

```