#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100010;
const int MAX_L = 18;
int father[MAX_L][MAX_N];
vector<int> adj[MAX_N];
int depth[MAX_N];
vector<int> visited;
int T = 0;
int IN[MAX_N], OUT[MAX_N];
vector<int> Perlevel[MAX_N];
void dfs(int u, int p = -1) {
  T++;
  IN[u] = T;
  visited[u] = true;
  Perlevel[depth[u]].emplace_back(T);
  for (auto v : adj[u]) {
    if (!visited[v]) {
      depth[v] = depth[u] + 1;
      father[0][v] = u;
      dfs(v, u);
    }
  }
  OUT[u] = T;
}
int getKth(int x, int k) {
  for (int i = MAX_L; i >= 0; i--) {
    if (k & (1 << i)) {
      x = father[i][x];
      if (x == -1) break;
    }
  }
  return x;
}
int main() {
  int n;
  cin >> n;
  visited.resize(n, false);
  memset(father, -1, sizeof(father));
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    x--;
    if (x > -1) {
      adj[x].emplace_back(i);
    }
  }
  for (int i = 0; i < n; i++) {
    if (!visited[i]) {
      dfs(i);
    }
  }
  for (int i = 1; i < MAX_L; i++) {
    for (int j = 0; j < n; j++) {
      int x = father[i - 1][j];
      if (x == -1) continue;
      father[i][j] = father[i - 1][x];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int x, k;
    cin >> x >> k;
    x--;
    int fatherN = getKth(x, k);
    if (fatherN == -1)
      cout << 0;
    else {
      int ans = upper_bound(Perlevel[depth[x]].begin(),
                            Perlevel[depth[x]].end(), OUT[fatherN]) -
                lower_bound(Perlevel[depth[x]].begin(),
                            Perlevel[depth[x]].end(), IN[fatherN]);
      cout << (ans - 1 < 0 ? 0 : ans - 1);
    }
    cout << endl;
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

Polycarpus got hold of a family relationship tree. The tree describes family relationships of n people, numbered 1 through n. Each person in the tree has no more than one parent.

Let's call person a a 1-ancestor of person b, if a is the parent of b.

Let's call person a a k-ancestor (k > 1) of person b, if person b has a 1-ancestor, and a is a (k - 1)-ancestor of b's 1-ancestor. 

Family relationships don't form cycles in the found tree. In other words, there is no person who is his own ancestor, directly or indirectly (that is, who is an x-ancestor for himself, for some x, x > 0).

Let's call two people x and y (x ≠ y) p-th cousins (p > 0), if there is person z, who is a p-ancestor of x and a p-ancestor of y.

Polycarpus wonders how many counsins and what kinds of them everybody has. He took a piece of paper and wrote m pairs of integers vi, pi. Help him to calculate the number of pi-th cousins that person vi has, for each pair vi, pi.

Input

The first input line contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. The next line contains n space-separated integers r1, r2, ..., rn, where ri (1 ≤ ri ≤ n) is the number of person i's parent or 0, if person i has no parent. It is guaranteed that family relationships don't form cycles.

The third line contains a single number m (1 ≤ m ≤ 105) — the number of family relationship queries Polycarus has. Next m lines contain pairs of space-separated integers. The i-th line contains numbers vi, pi (1 ≤ vi, pi ≤ n).

Output

Print m space-separated integers — the answers to Polycarpus' queries. Print the answers to the queries in the order, in which the queries occur in the input.

Examples

Input

6
0 1 1 0 4 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

0 0 1 0 0 1 1 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100010;
const int MAX_L = 18;
int father[MAX_L][MAX_N];
vector<int> adj[MAX_N];
int depth[MAX_N];
vector<int> visited;
int T = 0;
int IN[MAX_N], OUT[MAX_N];
vector<int> Perlevel[MAX_N];
void dfs(int u, int p = -1) {
  T++;
  IN[u] = T;
  visited[u] = true;
  Perlevel[depth[u]].emplace_back(T);
  for (auto v : adj[u]) {
    if (!visited[v]) {
      depth[v] = depth[u] + 1;
      father[0][v] = u;
      dfs(v, u);
    }
  }
  OUT[u] = T;
}
int getKth(int x, int k) {
  for (int i = MAX_L; i >= 0; i--) {
    if (k & (1 << i)) {
      x = father[i][x];
      if (x == -1) break;
    }
  }
  return x;
}
int main() {
  int n;
  cin >> n;
  visited.resize(n, false);
  memset(father, -1, sizeof(father));
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    x--;
    if (x > -1) {
      adj[x].emplace_back(i);
    }
  }
  for (int i = 0; i < n; i++) {
    if (!visited[i]) {
      dfs(i);
    }
  }
  for (int i = 1; i < MAX_L; i++) {
    for (int j = 0; j < n; j++) {
      int x = father[i - 1][j];
      if (x == -1) continue;
      father[i][j] = father[i - 1][x];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int x, k;
    cin >> x >> k;
    x--;
    int fatherN = getKth(x, k);
    if (fatherN == -1)
      cout << 0;
    else {
      int ans = upper_bound(Perlevel[depth[x]].begin(),
                            Perlevel[depth[x]].end(), OUT[fatherN]) -
                lower_bound(Perlevel[depth[x]].begin(),
                            Perlevel[depth[x]].end(), IN[fatherN]);
      cout << (ans - 1 < 0 ? 0 : ans - 1);
    }
    cout << endl;
  }
}

```