#include <bits/stdc++.h>
using namespace std;
const int N = 300004;
const int MOD = 1e9 + 7;
int n, q, czas, h[N], p[N], In[N], Out[N], treeX[4 * N], treeK[4 * N];
vector<int> G[N];
bool odw[N];
void DFS(int u) {
  odw[u] = true;
  In[u] = ++czas;
  for (auto v : G[u]) {
    if (odw[v] == false) {
      h[v] = h[u] + 1;
      DFS(v);
    }
  }
  Out[u] = czas;
}
void Insert(int v, int l, int r, int a, int b, int c1, int c2) {
  if (r < a || b < l || l > r) return;
  if (a <= l && r <= b) {
    treeX[v] += c1;
    treeX[v] %= MOD;
    treeK[v] += c2;
    treeK[v] %= MOD;
    return;
  }
  int mid = (l + r) / 2;
  Insert(2 * v, l, mid, a, b, c1, c2);
  Insert(2 * v + 1, mid + 1, r, a, b, c1, c2);
}
int QueryX(int v, int l, int r, int x) {
  if (r < x || x < l || l > r) return 0;
  if (x == l && x == r) return treeX[v];
  int mid = (l + r) / 2;
  if (l <= x && x <= mid)
    return (treeX[v] + QueryX(2 * v, l, mid, x)) % MOD;
  else if (mid + 1 <= x && x <= r)
    return (treeX[v] + QueryX(2 * v + 1, mid + 1, r, x)) % MOD;
}
int QueryK(int v, int l, int r, int x) {
  if (r < x || x < l || l > r) return 0;
  if (l == x && r == x) return treeK[v];
  int mid = (l + r) / 2;
  if (l <= x && x <= mid)
    return (treeK[v] + QueryK(2 * v, l, mid, x)) % MOD;
  else if (mid + 1 <= x && x <= r)
    return (treeK[v] + QueryK(2 * v + 1, mid + 1, r, x)) % MOD;
}
int normalizuj(long long x) {
  if (x < 0) {
    x = -x;
    x %= MOD;
    x = MOD - x;
  }
  return x %= MOD;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &p[i]);
    G[i].push_back(p[i]);
    G[p[i]].push_back(i);
  }
  DFS(1);
  scanf("%d", &q);
  while (q--) {
    int numer_zap;
    scanf("%d", &numer_zap);
    if (numer_zap == 1) {
      int u, x, k;
      scanf("%d%d%d", &u, &x, &k);
      Insert(1, 1, n, In[u], Out[u], ((long long)x + (long long)h[u] * k) % MOD,
             k);
    } else {
      int v;
      scanf("%d", &v);
      int wynik = normalizuj((long long)QueryX(1, 1, n, In[v]) -
                             (long long)h[v] * QueryK(1, 1, n, In[v]));
      printf("%d\n", wynik);
    }
  }
  getchar();
  getchar();
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 N = 300004;
const int MOD = 1e9 + 7;
int n, q, czas, h[N], p[N], In[N], Out[N], treeX[4 * N], treeK[4 * N];
vector<int> G[N];
bool odw[N];
void DFS(int u) {
  odw[u] = true;
  In[u] = ++czas;
  for (auto v : G[u]) {
    if (odw[v] == false) {
      h[v] = h[u] + 1;
      DFS(v);
    }
  }
  Out[u] = czas;
}
void Insert(int v, int l, int r, int a, int b, int c1, int c2) {
  if (r < a || b < l || l > r) return;
  if (a <= l && r <= b) {
    treeX[v] += c1;
    treeX[v] %= MOD;
    treeK[v] += c2;
    treeK[v] %= MOD;
    return;
  }
  int mid = (l + r) / 2;
  Insert(2 * v, l, mid, a, b, c1, c2);
  Insert(2 * v + 1, mid + 1, r, a, b, c1, c2);
}
int QueryX(int v, int l, int r, int x) {
  if (r < x || x < l || l > r) return 0;
  if (x == l && x == r) return treeX[v];
  int mid = (l + r) / 2;
  if (l <= x && x <= mid)
    return (treeX[v] + QueryX(2 * v, l, mid, x)) % MOD;
  else if (mid + 1 <= x && x <= r)
    return (treeX[v] + QueryX(2 * v + 1, mid + 1, r, x)) % MOD;
}
int QueryK(int v, int l, int r, int x) {
  if (r < x || x < l || l > r) return 0;
  if (l == x && r == x) return treeK[v];
  int mid = (l + r) / 2;
  if (l <= x && x <= mid)
    return (treeK[v] + QueryK(2 * v, l, mid, x)) % MOD;
  else if (mid + 1 <= x && x <= r)
    return (treeK[v] + QueryK(2 * v + 1, mid + 1, r, x)) % MOD;
}
int normalizuj(long long x) {
  if (x < 0) {
    x = -x;
    x %= MOD;
    x = MOD - x;
  }
  return x %= MOD;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &p[i]);
    G[i].push_back(p[i]);
    G[p[i]].push_back(i);
  }
  DFS(1);
  scanf("%d", &q);
  while (q--) {
    int numer_zap;
    scanf("%d", &numer_zap);
    if (numer_zap == 1) {
      int u, x, k;
      scanf("%d%d%d", &u, &x, &k);
      Insert(1, 1, n, In[u], Out[u], ((long long)x + (long long)h[u] * k) % MOD,
             k);
    } else {
      int v;
      scanf("%d", &v);
      int wynik = normalizuj((long long)QueryX(1, 1, n, In[v]) -
                             (long long)h[v] * QueryK(1, 1, n, In[v]));
      printf("%d\n", wynik);
    }
  }
  getchar();
  getchar();
  return 0;
}

```