#include <bits/stdc++.h>
using namespace std;
int n, m, tot;
int anc[100100][20];
int dep[100100];
int l[100100], r[100100];
vector<int> v[100100], u[100100];
void dfs(int x) {
  l[x] = ++tot;
  u[dep[x]].push_back(tot);
  for (auto y : v[x]) {
    dep[y] = dep[x] + 1;
    dfs(y);
  }
  r[x] = tot;
}
int query(int x, int p) {
  int d = dep[x];
  if (p >= d) return 0;
  for (int i = 19; i >= 0; i--) {
    if (p >= (1 << i)) {
      p = p - (1 << i);
      x = anc[x][i];
    }
  }
  int up = upper_bound(u[d].begin(), u[d].end(), r[x]) - u[d].begin();
  int lo = lower_bound(u[d].begin(), u[d].end(), l[x]) - u[d].begin();
  return up - lo - 1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    anc[i][0] = x;
    v[x].push_back(i);
  }
  dfs(0);
  for (int j = 1; j < 20; j++) {
    for (int i = 1; i <= n; i++) {
      int last = anc[i][j - 1];
      anc[i][j] = anc[last][j - 1];
    }
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    int ans = query(x, y);
    cout << ans << " ";
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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, m, tot;
int anc[100100][20];
int dep[100100];
int l[100100], r[100100];
vector<int> v[100100], u[100100];
void dfs(int x) {
  l[x] = ++tot;
  u[dep[x]].push_back(tot);
  for (auto y : v[x]) {
    dep[y] = dep[x] + 1;
    dfs(y);
  }
  r[x] = tot;
}
int query(int x, int p) {
  int d = dep[x];
  if (p >= d) return 0;
  for (int i = 19; i >= 0; i--) {
    if (p >= (1 << i)) {
      p = p - (1 << i);
      x = anc[x][i];
    }
  }
  int up = upper_bound(u[d].begin(), u[d].end(), r[x]) - u[d].begin();
  int lo = lower_bound(u[d].begin(), u[d].end(), l[x]) - u[d].begin();
  return up - lo - 1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    anc[i][0] = x;
    v[x].push_back(i);
  }
  dfs(0);
  for (int j = 1; j < 20; j++) {
    for (int i = 1; i <= n; i++) {
      int last = anc[i][j - 1];
      anc[i][j] = anc[last][j - 1];
    }
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    int ans = query(x, y);
    cout << ans << " ";
  }
}

```