#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> adj[200005];
int dep[200005];
int lca[200005][21];
int st[200005], ed[200005];
int cnt = 0;
vector<int> ht[200005];
void dfs(int root, int prev) {
  st[root] = cnt++;
  ht[dep[root]].push_back(st[root]);
  for (int ver : adj[root]) {
    if (ver != prev) {
      dep[ver] = dep[root] + 1;
      dfs(ver, root);
    }
  }
  ed[root] = cnt++;
}
void fillLCATable() {
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++) {
      if (lca[j][i - 1] != 0) {
        int par = lca[j][i - 1];
        lca[j][i] = lca[par][i - 1];
      }
    }
}
int walk(int x, int diff) {
  for (int i = 0; i < 20; i++) {
    if ((diff & 1)) x = lca[x][i];
    diff = (diff >> 1);
  }
  return x;
}
int main() {
  int m;
  cin >> n;
  for (int i = 0; i < 200005; i++)
    for (int j = 0; j <= 20; j++) lca[i][j] = 0;
  for (int i = 0; i < n; i++) {
    cin >> lca[i + 1][0];
    if (lca[i + 1][0] != 0) {
      adj[i + 1].push_back(lca[i + 1][0]);
      adj[lca[i + 1][0]].push_back(i + 1);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (lca[i][0] == 0) {
      dep[i] = 0;
      dfs(i, i);
    }
  }
  fillLCATable();
  cin >> m;
  int a, b;
  vector<int> res;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    if (b > dep[a]) {
      res.push_back(0);
      continue;
    }
    int prev_node = walk(a, b);
    int lval =
        lower_bound(ht[dep[a]].begin(), ht[dep[a]].end(), st[prev_node]) -
        ht[dep[a]].begin();
    int hval =
        upper_bound(ht[dep[a]].begin(), ht[dep[a]].end(), ed[prev_node]) -
        ht[dep[a]].begin();
    res.push_back(hval - lval - 1);
  }
  for (int i = 0; i < res.size(); i++) {
    cout << res[i] << " ";
  }
  cout << endl;
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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;
int n;
vector<int> adj[200005];
int dep[200005];
int lca[200005][21];
int st[200005], ed[200005];
int cnt = 0;
vector<int> ht[200005];
void dfs(int root, int prev) {
  st[root] = cnt++;
  ht[dep[root]].push_back(st[root]);
  for (int ver : adj[root]) {
    if (ver != prev) {
      dep[ver] = dep[root] + 1;
      dfs(ver, root);
    }
  }
  ed[root] = cnt++;
}
void fillLCATable() {
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++) {
      if (lca[j][i - 1] != 0) {
        int par = lca[j][i - 1];
        lca[j][i] = lca[par][i - 1];
      }
    }
}
int walk(int x, int diff) {
  for (int i = 0; i < 20; i++) {
    if ((diff & 1)) x = lca[x][i];
    diff = (diff >> 1);
  }
  return x;
}
int main() {
  int m;
  cin >> n;
  for (int i = 0; i < 200005; i++)
    for (int j = 0; j <= 20; j++) lca[i][j] = 0;
  for (int i = 0; i < n; i++) {
    cin >> lca[i + 1][0];
    if (lca[i + 1][0] != 0) {
      adj[i + 1].push_back(lca[i + 1][0]);
      adj[lca[i + 1][0]].push_back(i + 1);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (lca[i][0] == 0) {
      dep[i] = 0;
      dfs(i, i);
    }
  }
  fillLCATable();
  cin >> m;
  int a, b;
  vector<int> res;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    if (b > dep[a]) {
      res.push_back(0);
      continue;
    }
    int prev_node = walk(a, b);
    int lval =
        lower_bound(ht[dep[a]].begin(), ht[dep[a]].end(), st[prev_node]) -
        ht[dep[a]].begin();
    int hval =
        upper_bound(ht[dep[a]].begin(), ht[dep[a]].end(), ed[prev_node]) -
        ht[dep[a]].begin();
    res.push_back(hval - lval - 1);
  }
  for (int i = 0; i < res.size(); i++) {
    cout << res[i] << " ";
  }
  cout << endl;
  return 0;
}

```