#include <bits/stdc++.h>
using namespace std;
long long int modulo = 1e9 + 7;
const int limite = 300001;
void insertar(int primero, vector<long long int> &acumx,
              vector<long long int> &acumk, vector<int> &cuantos, int pos,
              int x, int k) {
  pos += primero;
  cuantos[pos] = 1;
  acumx[pos] += x;
  acumk[pos] += k;
  acumx[pos] %= modulo;
  acumk[pos] %= modulo;
  while (pos > 1) {
    pos /= 2;
    cuantos[pos] = cuantos[2 * pos] + cuantos[2 * pos + 1];
    acumx[pos] = (acumx[2 * pos + 1] + acumx[2 * pos] -
                  acumk[2 * pos] * cuantos[2 * pos + 1]) %
                 modulo;
    acumk[pos] = (acumk[2 * pos] + acumk[2 * pos + 1]) % modulo;
  }
}
long long int calcula(int primero, vector<long long int> &acumx,
                      vector<long long int> &acumk, vector<int> &cuantos,
                      int pos, long long int &cuan) {
  pos += primero;
  long long int x = acumx[pos] - acumk[pos] * cuan;
  cuan++;
  while (pos > 1) {
    if (pos % 2 == 1) {
      x = (x + acumx[pos - 1] - acumk[pos - 1] * cuan);
      cuan += cuantos[pos - 1];
    }
    pos /= 2;
  }
  return x;
}
vector<int> g[limite];
int padre[limite];
int nodos[limite];
int primero[limite];
int tope[limite];
vector<long long int> acumx[limite];
vector<long long int> acumk[limite];
vector<int> cuantos[limite];
int superpadre[limite];
int superhermano[limite];
int superposicion[limite];
void insertar(int u, int x, int k) {
  int her = superhermano[u];
  int pos = superposicion[u];
  insertar(primero[her], acumx[her], acumk[her], cuantos[her], pos, x, k);
}
long long int calcula(int u) {
  long long int x = 0;
  long long int cuan = 0;
  for (;;) {
    int her = superhermano[u];
    int suppos = superposicion[u];
    x += calcula(primero[her], acumx[her], acumk[her], cuantos[her], suppos,
                 cuan);
    if (her == 1) break;
    u = superpadre[u];
  }
  return x;
}
int computanodos(int u) {
  nodos[u] = 1;
  vector<int> &ar = g[u];
  for (int i = 0; i < int(ar.size()); i++) nodos[u] += computanodos(ar[i]);
  return nodos[u];
}
void genera(int u, int sup) {
  vector<int> heavy;
  for (;;) {
    int unext = u;
    heavy.push_back(u);
    vector<int> &ar = g[u];
    for (int i = 0; i < int(ar.size()); i++) {
      int v = ar[i];
      if (2 * nodos[v] > nodos[u]) {
        unext = v;
      } else {
        genera(v, u);
      }
    }
    if (u == unext) break;
    u = unext;
  }
  int her = heavy[0];
  for (int i = 0; i < int(heavy.size()); i++) {
    u = heavy[i];
    superpadre[u] = sup;
    superhermano[u] = her;
    superposicion[u] = i;
  }
  {
    int nn = int(heavy.size()) + 1;
    primero[her] = 1;
    while (nn > 0) {
      nn /= 2;
      primero[her] *= 2;
    }
    tope[her] = primero[her] * 2;
    acumx[her] = acumk[her] = vector<long long int>(tope[her], 0);
    cuantos[her] = vector<int>(tope[her], 0);
    for (int i = 0; i < int(heavy.size()); i++)
      insertar(primero[her], acumx[her], acumk[her], cuantos[her], i, 0, 0);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int u = 2; u <= n; u++) {
    cin >> padre[u];
    g[padre[u]].push_back(u);
  }
  computanodos(1);
  genera(1, 0);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int tipo;
    cin >> tipo;
    if (tipo == 1) {
      int u;
      long long int x, k;
      cin >> u >> x >> k;
      insertar(u, x, k);
    } else {
      int u;
      cin >> u;
      cout << (calcula(u) % modulo + modulo) % modulo << endl;
    }
  }
}

### 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;
long long int modulo = 1e9 + 7;
const int limite = 300001;
void insertar(int primero, vector<long long int> &acumx,
              vector<long long int> &acumk, vector<int> &cuantos, int pos,
              int x, int k) {
  pos += primero;
  cuantos[pos] = 1;
  acumx[pos] += x;
  acumk[pos] += k;
  acumx[pos] %= modulo;
  acumk[pos] %= modulo;
  while (pos > 1) {
    pos /= 2;
    cuantos[pos] = cuantos[2 * pos] + cuantos[2 * pos + 1];
    acumx[pos] = (acumx[2 * pos + 1] + acumx[2 * pos] -
                  acumk[2 * pos] * cuantos[2 * pos + 1]) %
                 modulo;
    acumk[pos] = (acumk[2 * pos] + acumk[2 * pos + 1]) % modulo;
  }
}
long long int calcula(int primero, vector<long long int> &acumx,
                      vector<long long int> &acumk, vector<int> &cuantos,
                      int pos, long long int &cuan) {
  pos += primero;
  long long int x = acumx[pos] - acumk[pos] * cuan;
  cuan++;
  while (pos > 1) {
    if (pos % 2 == 1) {
      x = (x + acumx[pos - 1] - acumk[pos - 1] * cuan);
      cuan += cuantos[pos - 1];
    }
    pos /= 2;
  }
  return x;
}
vector<int> g[limite];
int padre[limite];
int nodos[limite];
int primero[limite];
int tope[limite];
vector<long long int> acumx[limite];
vector<long long int> acumk[limite];
vector<int> cuantos[limite];
int superpadre[limite];
int superhermano[limite];
int superposicion[limite];
void insertar(int u, int x, int k) {
  int her = superhermano[u];
  int pos = superposicion[u];
  insertar(primero[her], acumx[her], acumk[her], cuantos[her], pos, x, k);
}
long long int calcula(int u) {
  long long int x = 0;
  long long int cuan = 0;
  for (;;) {
    int her = superhermano[u];
    int suppos = superposicion[u];
    x += calcula(primero[her], acumx[her], acumk[her], cuantos[her], suppos,
                 cuan);
    if (her == 1) break;
    u = superpadre[u];
  }
  return x;
}
int computanodos(int u) {
  nodos[u] = 1;
  vector<int> &ar = g[u];
  for (int i = 0; i < int(ar.size()); i++) nodos[u] += computanodos(ar[i]);
  return nodos[u];
}
void genera(int u, int sup) {
  vector<int> heavy;
  for (;;) {
    int unext = u;
    heavy.push_back(u);
    vector<int> &ar = g[u];
    for (int i = 0; i < int(ar.size()); i++) {
      int v = ar[i];
      if (2 * nodos[v] > nodos[u]) {
        unext = v;
      } else {
        genera(v, u);
      }
    }
    if (u == unext) break;
    u = unext;
  }
  int her = heavy[0];
  for (int i = 0; i < int(heavy.size()); i++) {
    u = heavy[i];
    superpadre[u] = sup;
    superhermano[u] = her;
    superposicion[u] = i;
  }
  {
    int nn = int(heavy.size()) + 1;
    primero[her] = 1;
    while (nn > 0) {
      nn /= 2;
      primero[her] *= 2;
    }
    tope[her] = primero[her] * 2;
    acumx[her] = acumk[her] = vector<long long int>(tope[her], 0);
    cuantos[her] = vector<int>(tope[her], 0);
    for (int i = 0; i < int(heavy.size()); i++)
      insertar(primero[her], acumx[her], acumk[her], cuantos[her], i, 0, 0);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int u = 2; u <= n; u++) {
    cin >> padre[u];
    g[padre[u]].push_back(u);
  }
  computanodos(1);
  genera(1, 0);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int tipo;
    cin >> tipo;
    if (tipo == 1) {
      int u;
      long long int x, k;
      cin >> u >> x >> k;
      insertar(u, x, k);
    } else {
      int u;
      cin >> u;
      cout << (calcula(u) % modulo + modulo) % modulo << endl;
    }
  }
}

```