#include <bits/stdc++.h>
using namespace std;
const int maxN = 100 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int maxL = 20;
vector<int> a[maxN], xx[maxN];
int st[maxN], en[maxN], h[maxN], p[maxN];
int lca[maxN][maxL], cnt;
void dfs(int v, int par = 0) {
  st[v] = cnt++;
  lca[v][0] = par;
  for (int i = 1; i < maxL; i++) lca[v][i] = lca[lca[v][i - 1]][i - 1];
  xx[h[v]].push_back(st[v]);
  for (auto u : a[v])
    if (u - par) h[u] = h[v] + 1, dfs(u, v);
  en[v] = cnt;
}
int get_par(int v, int k) {
  for (int i = 0; i < maxL; i++)
    if (k >> i & 1) v = lca[v][i];
  return v;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    cin >> p[i];
    if (p[i]) a[p[i]].push_back(i);
  }
  for (int i = 1; i < n + 1; i++)
    if (p[i] == 0) dfs(i);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int v, k;
    cin >> v >> k;
    int dis = h[v];
    if (k > dis) {
      cout << 0 << ' ';
      continue;
    }
    v = get_par(v, k);
    int psl =
        lower_bound(xx[dis].begin(), xx[dis].end(), st[v]) - xx[dis].begin();
    int psr =
        lower_bound(xx[dis].begin(), xx[dis].end(), en[v]) - xx[dis].begin();
    cout << psr - psl - 1 << ' ';
  }
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int maxN = 100 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int maxL = 20;
vector<int> a[maxN], xx[maxN];
int st[maxN], en[maxN], h[maxN], p[maxN];
int lca[maxN][maxL], cnt;
void dfs(int v, int par = 0) {
  st[v] = cnt++;
  lca[v][0] = par;
  for (int i = 1; i < maxL; i++) lca[v][i] = lca[lca[v][i - 1]][i - 1];
  xx[h[v]].push_back(st[v]);
  for (auto u : a[v])
    if (u - par) h[u] = h[v] + 1, dfs(u, v);
  en[v] = cnt;
}
int get_par(int v, int k) {
  for (int i = 0; i < maxL; i++)
    if (k >> i & 1) v = lca[v][i];
  return v;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    cin >> p[i];
    if (p[i]) a[p[i]].push_back(i);
  }
  for (int i = 1; i < n + 1; i++)
    if (p[i] == 0) dfs(i);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int v, k;
    cin >> v >> k;
    int dis = h[v];
    if (k > dis) {
      cout << 0 << ' ';
      continue;
    }
    v = get_par(v, k);
    int psl =
        lower_bound(xx[dis].begin(), xx[dis].end(), st[v]) - xx[dis].begin();
    int psr =
        lower_bound(xx[dis].begin(), xx[dis].end(), en[v]) - xx[dis].begin();
    cout << psr - psl - 1 << ' ';
  }
  cout << endl;
  return 0;
}

```