#include <bits/stdc++.h>
using namespace std;
std::vector<int> adj[100010], level[100010];
int v[100010], par[100010], Level[100010], t = 0, dis[100010], fin[100010];
void dfs(int s, int l) {
  if (v[s]) return;
  v[s] = 1;
  dis[s] = t++;
  for (int i = 0; i < adj[s].size(); ++i) {
    int t = adj[s][i];
    if (!v[t]) {
      par[t] = s;
      dfs(t, l + 1);
    }
  }
  level[l].push_back(t);
  fin[s] = t++;
  Level[s] = l;
}
int anch[100010][35];
void sparse_table(int n) {
  for (int i = 0; (1 << i) <= n; ++i)
    for (int j = 1; j <= n; ++j) {
      if (!i)
        anch[j][i] = par[j];
      else if (anch[j][i - 1] != -1)
        anch[j][i] = anch[anch[j][i - 1]][i - 1];
    }
}
int parent(int x, int p) {
  for (int i = 30; i >= 0; i--) {
    if ((p >= (1 << i)) and anch[x][i] != -1) {
      p -= (1 << i);
      x = anch[x][i];
    }
  }
  return x;
}
int main() {
  memset(anch, -1, sizeof anch);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    if (x) adj[x].push_back(i);
  }
  for (int i = 1; i <= n; ++i)
    if (!v[i]) par[i] = -1, dfs(i, 0);
  sparse_table(n);
  int q;
  cin >> q;
  while (q--) {
    int p, x;
    cin >> x >> p;
    if (Level[x] < p or par[x] == -1)
      printf("0 ");
    else {
      int l = Level[x];
      int P = parent(x, p);
      int up = upper_bound(level[l].begin(), level[l].end(), fin[P]) -
               level[l].begin();
      int lo = lower_bound(level[l].begin(), level[l].end(), dis[P]) -
               level[l].begin();
      printf("%d ", up - lo - 1);
    }
  }
}

### 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;
std::vector<int> adj[100010], level[100010];
int v[100010], par[100010], Level[100010], t = 0, dis[100010], fin[100010];
void dfs(int s, int l) {
  if (v[s]) return;
  v[s] = 1;
  dis[s] = t++;
  for (int i = 0; i < adj[s].size(); ++i) {
    int t = adj[s][i];
    if (!v[t]) {
      par[t] = s;
      dfs(t, l + 1);
    }
  }
  level[l].push_back(t);
  fin[s] = t++;
  Level[s] = l;
}
int anch[100010][35];
void sparse_table(int n) {
  for (int i = 0; (1 << i) <= n; ++i)
    for (int j = 1; j <= n; ++j) {
      if (!i)
        anch[j][i] = par[j];
      else if (anch[j][i - 1] != -1)
        anch[j][i] = anch[anch[j][i - 1]][i - 1];
    }
}
int parent(int x, int p) {
  for (int i = 30; i >= 0; i--) {
    if ((p >= (1 << i)) and anch[x][i] != -1) {
      p -= (1 << i);
      x = anch[x][i];
    }
  }
  return x;
}
int main() {
  memset(anch, -1, sizeof anch);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    if (x) adj[x].push_back(i);
  }
  for (int i = 1; i <= n; ++i)
    if (!v[i]) par[i] = -1, dfs(i, 0);
  sparse_table(n);
  int q;
  cin >> q;
  while (q--) {
    int p, x;
    cin >> x >> p;
    if (Level[x] < p or par[x] == -1)
      printf("0 ");
    else {
      int l = Level[x];
      int P = parent(x, p);
      int up = upper_bound(level[l].begin(), level[l].end(), fin[P]) -
               level[l].begin();
      int lo = lower_bound(level[l].begin(), level[l].end(), dis[P]) -
               level[l].begin();
      printf("%d ", up - lo - 1);
    }
  }
}

```