#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5 + 5, mod = 1e9 + 7;
long long n, m, tot, head[N], fa[N], siz[N], son[N], dep[N], dfn[N], tim,
    top[N];
struct node {
  long long to, nex;
} e[N];
struct tree {
  long long l, r, tag, val;
} tr[N << 2];
inline void add(long long x, long long y) {
  e[++tot] = (node){y, head[x]};
  head[x] = tot;
}
inline void dfs(long long x, long long f) {
  dep[x] = dep[f] + 1;
  siz[x] = 1;
  fa[x] = f;
  long long k;
  for (register long long i = head[x]; i; i = e[i].nex) {
    k = e[i].to;
    dfs(k, x);
    siz[x] += siz[k];
    if (siz[son[x]] < siz[k]) son[x] = k;
  }
}
inline void dfs2(long long x, long long f) {
  dfn[x] = ++tim;
  top[x] = f;
  if (!son[x]) return;
  dfs2(son[x], f);
  long long k;
  for (register long long i = head[x]; i; i = e[i].nex) {
    k = e[i].to;
    if (k == son[x]) continue;
    dfs2(k, k);
  }
}
inline void build(long long k, long long l, long long r) {
  tr[k].l = l;
  tr[k].r = r;
  if (l == r) return;
  long long mid = (l + r) / 2;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
}
inline void update(long long k) {
  if (!tr[k].tag) return;
  tr[k << 1].val =
      (tr[k << 1].val + tr[k].tag * (tr[k << 1].r - tr[k << 1].l + 1) % mod) %
      mod;
  tr[k << 1 | 1].val =
      (tr[k << 1 | 1].val +
       tr[k].tag * (tr[k << 1 | 1].r - tr[k << 1 | 1].l + 1) % mod) %
      mod;
  tr[k << 1].tag = (tr[k << 1].tag + tr[k].tag) % mod;
  tr[k << 1 | 1].tag = (tr[k << 1 | 1].tag + tr[k].tag) % mod;
  tr[k].tag = 0;
}
inline void change(long long k, long long st, long long ed, long long v) {
  if (tr[k].l > ed || tr[k].r < st) return;
  if (tr[k].l >= st && tr[k].r <= ed) {
    tr[k].val = (tr[k].val + v * (tr[k].r - tr[k].l + 1) % mod) % mod;
    tr[k].tag = (tr[k].tag + v) % mod;
    return;
  }
  update(k);
  change(k << 1, st, ed, v);
  change(k << 1 | 1, st, ed, v);
  tr[k].val = (tr[k << 1].val + tr[k << 1 | 1].val) % mod;
}
inline long long query(long long k, long long st, long long ed) {
  if (tr[k].l > ed || tr[k].r < st) return 0;
  if (tr[k].l >= st && tr[k].r <= ed) return tr[k].val;
  update(k);
  return (query(k << 1, st, ed) + query(k << 1 | 1, st, ed)) % mod;
}
inline long long Ask(long long x, long long y) {
  long long res = 0;
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    res = (res + query(1, dfn[top[x]], dfn[x])) % mod;
    x = fa[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  res = (res + query(1, dfn[x], dfn[y])) % mod;
  return (res % mod + mod) % mod;
}
void read(long long &x) {
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') s = getchar();
  while (s <= '9' && s >= '0') {
    x = x * 10 + s - '0';
    s = getchar();
  }
}
void print(long long x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
signed main() {
  read(n);
  long long u;
  for (register long long i = 2; i <= n; i++) {
    read(u);
    add(u, i);
  }
  dfs(1, 0);
  dfs2(1, 1);
  build(1, 1, n);
  read(m);
  long long opt, v, w;
  while (m--) {
    read(opt);
    read(u);
    if (opt == 1) {
      read(v);
      read(w);
      change(1, dfn[u], dfn[u], v);
      change(1, dfn[u] + 1, dfn[u] + siz[u] - 1, mod - w);
    } else {
      print(Ask(1, u));
      putchar('\n');
    }
  }
  return 0;
}

### Prompt

Please formulate 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 N = 3e5 + 5, mod = 1e9 + 7;
long long n, m, tot, head[N], fa[N], siz[N], son[N], dep[N], dfn[N], tim,
    top[N];
struct node {
  long long to, nex;
} e[N];
struct tree {
  long long l, r, tag, val;
} tr[N << 2];
inline void add(long long x, long long y) {
  e[++tot] = (node){y, head[x]};
  head[x] = tot;
}
inline void dfs(long long x, long long f) {
  dep[x] = dep[f] + 1;
  siz[x] = 1;
  fa[x] = f;
  long long k;
  for (register long long i = head[x]; i; i = e[i].nex) {
    k = e[i].to;
    dfs(k, x);
    siz[x] += siz[k];
    if (siz[son[x]] < siz[k]) son[x] = k;
  }
}
inline void dfs2(long long x, long long f) {
  dfn[x] = ++tim;
  top[x] = f;
  if (!son[x]) return;
  dfs2(son[x], f);
  long long k;
  for (register long long i = head[x]; i; i = e[i].nex) {
    k = e[i].to;
    if (k == son[x]) continue;
    dfs2(k, k);
  }
}
inline void build(long long k, long long l, long long r) {
  tr[k].l = l;
  tr[k].r = r;
  if (l == r) return;
  long long mid = (l + r) / 2;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
}
inline void update(long long k) {
  if (!tr[k].tag) return;
  tr[k << 1].val =
      (tr[k << 1].val + tr[k].tag * (tr[k << 1].r - tr[k << 1].l + 1) % mod) %
      mod;
  tr[k << 1 | 1].val =
      (tr[k << 1 | 1].val +
       tr[k].tag * (tr[k << 1 | 1].r - tr[k << 1 | 1].l + 1) % mod) %
      mod;
  tr[k << 1].tag = (tr[k << 1].tag + tr[k].tag) % mod;
  tr[k << 1 | 1].tag = (tr[k << 1 | 1].tag + tr[k].tag) % mod;
  tr[k].tag = 0;
}
inline void change(long long k, long long st, long long ed, long long v) {
  if (tr[k].l > ed || tr[k].r < st) return;
  if (tr[k].l >= st && tr[k].r <= ed) {
    tr[k].val = (tr[k].val + v * (tr[k].r - tr[k].l + 1) % mod) % mod;
    tr[k].tag = (tr[k].tag + v) % mod;
    return;
  }
  update(k);
  change(k << 1, st, ed, v);
  change(k << 1 | 1, st, ed, v);
  tr[k].val = (tr[k << 1].val + tr[k << 1 | 1].val) % mod;
}
inline long long query(long long k, long long st, long long ed) {
  if (tr[k].l > ed || tr[k].r < st) return 0;
  if (tr[k].l >= st && tr[k].r <= ed) return tr[k].val;
  update(k);
  return (query(k << 1, st, ed) + query(k << 1 | 1, st, ed)) % mod;
}
inline long long Ask(long long x, long long y) {
  long long res = 0;
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    res = (res + query(1, dfn[top[x]], dfn[x])) % mod;
    x = fa[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  res = (res + query(1, dfn[x], dfn[y])) % mod;
  return (res % mod + mod) % mod;
}
void read(long long &x) {
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') s = getchar();
  while (s <= '9' && s >= '0') {
    x = x * 10 + s - '0';
    s = getchar();
  }
}
void print(long long x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
signed main() {
  read(n);
  long long u;
  for (register long long i = 2; i <= n; i++) {
    read(u);
    add(u, i);
  }
  dfs(1, 0);
  dfs2(1, 1);
  build(1, 1, n);
  read(m);
  long long opt, v, w;
  while (m--) {
    read(opt);
    read(u);
    if (opt == 1) {
      read(v);
      read(w);
      change(1, dfn[u], dfn[u], v);
      change(1, dfn[u] + 1, dfn[u] + siz[u] - 1, mod - w);
    } else {
      print(Ask(1, u));
      putchar('\n');
    }
  }
  return 0;
}

```