#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000 * 100 + 5;
vector<int> adj[MAXN], layer[MAXN];
int depth[MAXN], s_time[MAXN], f_time[MAXN];
int not_root[MAXN];
int parent[MAXN][18];
int dfs_time;
void dfs(int v);
int find(int v, int p);
int solve(int l, int s);
int main() {
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int par;
    cin >> par;
    if (par != 0) {
      adj[par].push_back(i);
      not_root[i] = true;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!not_root[i]) dfs(i);
  for (int i = 1; i < 18; i++)
    for (int v = 1; v <= n; v++) parent[v][i] = parent[parent[v][i - 1]][i - 1];
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, p;
    cin >> v >> p;
    v = find(v, p);
    if (v == 0 || layer[depth[v] + p].size() == 0) {
      cout << 0 << ' ';
      continue;
    }
    int l = depth[v] + p;
    cout << upper_bound(layer[l].begin(), layer[l].end(), f_time[v]) -
                lower_bound(layer[l].begin(), layer[l].end(), s_time[v]) - 1
         << ' ';
  }
  cout << endl;
  return 0;
}
int solve(int l, int s) {
  int start, end, mid;
  start = 0;
  end = layer[l].size();
  while (start != end - 1) {
    mid = (start + end) / 2;
    if (layer[l][mid] < s)
      start = mid;
    else
      end = mid;
  }
  return start;
}
int find(int v, int p) {
  for (int i = 17; i >= 0; i--)
    if ((1 << i) & p) v = parent[v][i];
  return v;
}
void dfs(int v) {
  s_time[v] = dfs_time;
  layer[depth[v]].push_back(s_time[v]);
  for (int i = 0; i < adj[v].size(); i++) {
    dfs_time++;
    int u = adj[v][i];
    depth[u] = depth[v] + 1;
    parent[u][0] = v;
    dfs(u);
  }
  f_time[v] = dfs_time++;
}

### 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 MAXN = 1000 * 100 + 5;
vector<int> adj[MAXN], layer[MAXN];
int depth[MAXN], s_time[MAXN], f_time[MAXN];
int not_root[MAXN];
int parent[MAXN][18];
int dfs_time;
void dfs(int v);
int find(int v, int p);
int solve(int l, int s);
int main() {
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int par;
    cin >> par;
    if (par != 0) {
      adj[par].push_back(i);
      not_root[i] = true;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!not_root[i]) dfs(i);
  for (int i = 1; i < 18; i++)
    for (int v = 1; v <= n; v++) parent[v][i] = parent[parent[v][i - 1]][i - 1];
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, p;
    cin >> v >> p;
    v = find(v, p);
    if (v == 0 || layer[depth[v] + p].size() == 0) {
      cout << 0 << ' ';
      continue;
    }
    int l = depth[v] + p;
    cout << upper_bound(layer[l].begin(), layer[l].end(), f_time[v]) -
                lower_bound(layer[l].begin(), layer[l].end(), s_time[v]) - 1
         << ' ';
  }
  cout << endl;
  return 0;
}
int solve(int l, int s) {
  int start, end, mid;
  start = 0;
  end = layer[l].size();
  while (start != end - 1) {
    mid = (start + end) / 2;
    if (layer[l][mid] < s)
      start = mid;
    else
      end = mid;
  }
  return start;
}
int find(int v, int p) {
  for (int i = 17; i >= 0; i--)
    if ((1 << i) & p) v = parent[v][i];
  return v;
}
void dfs(int v) {
  s_time[v] = dfs_time;
  layer[depth[v]].push_back(s_time[v]);
  for (int i = 0; i < adj[v].size(); i++) {
    dfs_time++;
    int u = adj[v][i];
    depth[u] = depth[v] + 1;
    parent[u][0] = v;
    dfs(u);
  }
  f_time[v] = dfs_time++;
}

```