#include <bits/stdc++.h>
using namespace std;
int n, l[100001], newn[100001], idx = -1, start, cnt[100001], t[100001],
                                p[100001][18];
vector<vector<int> > v(100001);
vector<int> roots;
bool visited[100001];
vector<int> L[100001];
int dfs(int node, int level, int parent) {
  if (visited[node]) return 0;
  visited[node] = 1;
  idx++;
  newn[node] = idx;
  l[newn[node]] = level;
  L[l[newn[node]]].push_back(newn[node]);
  t[newn[node]] = parent;
  int c = 1;
  for (int i = 0; i < v[node].size(); i++)
    c += dfs(v[node][i], level + 1, newn[node]);
  return cnt[newn[node]] = c;
}
void build() {
  for (int i = 0; i < n; i++) p[i][0] = t[i];
  for (int j = 1; 1 << j <= n; j++)
    for (int i = 0; i < n; i++) p[i][j] = p[p[i][j - 1]][j - 1];
}
int get(int a, int P) {
  int want = l[a] - P;
  for (int j = 17; j >= 0; j--)
    if (l[a] - (1 << j) >= want) a = p[a][j];
  return a;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    a--;
    if (a == -1) {
      roots.push_back(i);
      continue;
    }
    v[a].push_back(i);
    v[i].push_back(a);
  }
  for (int i = 0; i < roots.size(); i++)
    if (!visited[roots[i]]) start = idx + 1, dfs(roots[i], 0, roots[i]);
  build();
  for (int i = 0; i < 100001; i++) sort(L[i].begin(), L[i].end());
  int q;
  cin >> q;
  while (q--) {
    int v, P;
    cin >> v >> P;
    v--;
    v = newn[v];
    if (l[v] - P < 0) {
      cout << 0 << '\n';
      continue;
    }
    int u = get(v, P);
    int desiredlevel = l[u] + P;
    cout << upper_bound(L[desiredlevel].begin(), L[desiredlevel].end(),
                        u + cnt[u] - 1) -
                lower_bound(L[desiredlevel].begin(), L[desiredlevel].end(), u) -
                1
         << '\n';
  }
  return 0;
}

### Prompt

Generate a Cpp solution to 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, l[100001], newn[100001], idx = -1, start, cnt[100001], t[100001],
                                p[100001][18];
vector<vector<int> > v(100001);
vector<int> roots;
bool visited[100001];
vector<int> L[100001];
int dfs(int node, int level, int parent) {
  if (visited[node]) return 0;
  visited[node] = 1;
  idx++;
  newn[node] = idx;
  l[newn[node]] = level;
  L[l[newn[node]]].push_back(newn[node]);
  t[newn[node]] = parent;
  int c = 1;
  for (int i = 0; i < v[node].size(); i++)
    c += dfs(v[node][i], level + 1, newn[node]);
  return cnt[newn[node]] = c;
}
void build() {
  for (int i = 0; i < n; i++) p[i][0] = t[i];
  for (int j = 1; 1 << j <= n; j++)
    for (int i = 0; i < n; i++) p[i][j] = p[p[i][j - 1]][j - 1];
}
int get(int a, int P) {
  int want = l[a] - P;
  for (int j = 17; j >= 0; j--)
    if (l[a] - (1 << j) >= want) a = p[a][j];
  return a;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    a--;
    if (a == -1) {
      roots.push_back(i);
      continue;
    }
    v[a].push_back(i);
    v[i].push_back(a);
  }
  for (int i = 0; i < roots.size(); i++)
    if (!visited[roots[i]]) start = idx + 1, dfs(roots[i], 0, roots[i]);
  build();
  for (int i = 0; i < 100001; i++) sort(L[i].begin(), L[i].end());
  int q;
  cin >> q;
  while (q--) {
    int v, P;
    cin >> v >> P;
    v--;
    v = newn[v];
    if (l[v] - P < 0) {
      cout << 0 << '\n';
      continue;
    }
    int u = get(v, P);
    int desiredlevel = l[u] + P;
    cout << upper_bound(L[desiredlevel].begin(), L[desiredlevel].end(),
                        u + cnt[u] - 1) -
                lower_bound(L[desiredlevel].begin(), L[desiredlevel].end(), u) -
                1
         << '\n';
  }
  return 0;
}

```