#include <bits/stdc++.h>
using namespace std;
const int Maxn = 1e5 + 50;
int n, m, mark[Maxn], h[Maxn], st[Maxn], fi[Maxn], t, a[Maxn];
vector<int> y[Maxn];
vector<int> ert[Maxn];
vector<int> ans;
vector<int> test;
void main_dfs(int v, int q) {
  st[v] = t;
  ert[q].push_back(v);
  a[v] = (int)ert[q].size() - 1;
  h[v] = q;
  mark[v]++;
  for (auto u : y[v]) {
    if (mark[u] == 0) {
      t++;
      main_dfs(u, q + 1);
    }
  }
  fi[v] = t;
}
int find_parent(int v, int k) {
  int mh = h[v] - k;
  if (mh <= 0) {
    return 0;
  }
  if ((int)ert[mh].size() == 1) {
    return ert[mh][0];
  }
  int l = -1;
  int r = (int)ert[mh].size();
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (st[ert[mh][mid]] > fi[v]) {
      r = mid;
    } else {
      l = mid;
    }
  }
  return ert[mh][l];
}
int cnt(int v, int par) {
  if ((int)ert[h[v]].size() == 1) {
    return 0;
  }
  int l = a[v] - 1;
  int r = ert[h[v]].size() + 1;
  while (r - l > 1) {
    int mid = (r + l) / 2;
    if (st[ert[h[v]][mid]] > fi[par]) {
      r = mid;
    } else {
      l = mid;
    }
  }
  int end = l;
  l = -1;
  r = a[v] + 1;
  while (r - l > 1) {
    int mid = (r + l) / 2;
    if (fi[ert[h[v]][mid]] < st[par]) {
      l = mid;
    } else {
      r = mid;
    }
  }
  int start = r;
  return end - start;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int u;
    cin >> u;
    y[i].push_back(u);
    y[u].push_back(i);
  }
  main_dfs(0, 0);
  for (int i = 0; i < Maxn - 5; i++) {
    ert[i].push_back(Maxn - 5);
  }
  st[Maxn - 5] = 1e6;
  fi[Maxn - 5] = 1e6;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int v, p;
    cin >> v >> p;
    int vp = find_parent(v, p);
    if (vp == 0) {
      ans.push_back(0);
      continue;
    }
    ans.push_back(cnt(v, vp));
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << " ";
  }
}

### Prompt

Create a solution in cpp for 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 = 1e5 + 50;
int n, m, mark[Maxn], h[Maxn], st[Maxn], fi[Maxn], t, a[Maxn];
vector<int> y[Maxn];
vector<int> ert[Maxn];
vector<int> ans;
vector<int> test;
void main_dfs(int v, int q) {
  st[v] = t;
  ert[q].push_back(v);
  a[v] = (int)ert[q].size() - 1;
  h[v] = q;
  mark[v]++;
  for (auto u : y[v]) {
    if (mark[u] == 0) {
      t++;
      main_dfs(u, q + 1);
    }
  }
  fi[v] = t;
}
int find_parent(int v, int k) {
  int mh = h[v] - k;
  if (mh <= 0) {
    return 0;
  }
  if ((int)ert[mh].size() == 1) {
    return ert[mh][0];
  }
  int l = -1;
  int r = (int)ert[mh].size();
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (st[ert[mh][mid]] > fi[v]) {
      r = mid;
    } else {
      l = mid;
    }
  }
  return ert[mh][l];
}
int cnt(int v, int par) {
  if ((int)ert[h[v]].size() == 1) {
    return 0;
  }
  int l = a[v] - 1;
  int r = ert[h[v]].size() + 1;
  while (r - l > 1) {
    int mid = (r + l) / 2;
    if (st[ert[h[v]][mid]] > fi[par]) {
      r = mid;
    } else {
      l = mid;
    }
  }
  int end = l;
  l = -1;
  r = a[v] + 1;
  while (r - l > 1) {
    int mid = (r + l) / 2;
    if (fi[ert[h[v]][mid]] < st[par]) {
      l = mid;
    } else {
      r = mid;
    }
  }
  int start = r;
  return end - start;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int u;
    cin >> u;
    y[i].push_back(u);
    y[u].push_back(i);
  }
  main_dfs(0, 0);
  for (int i = 0; i < Maxn - 5; i++) {
    ert[i].push_back(Maxn - 5);
  }
  st[Maxn - 5] = 1e6;
  fi[Maxn - 5] = 1e6;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int v, p;
    cin >> v >> p;
    int vp = find_parent(v, p);
    if (vp == 0) {
      ans.push_back(0);
      continue;
    }
    ans.push_back(cnt(v, vp));
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << " ";
  }
}

```