#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> g(1e5 + 1);
int n;
vector<int> H(1e5 + 1);
vector<vector<int>> p(1e5 + 1, vector<int>(18, 0));
vector<vector<int>> R(1e5 + 1);
vector<int> tin(1e5 + 1);
vector<int> tout(1e5 + 1);
vector<int> num(1e5 + 1);
int cur = 0;
void dfs(int v, int par, int h = 0) {
  H[v] = h;
  R[h].push_back(v);
  num[v] = R[h].size() - 1;
  tin[v] = cur++;
  if (h > 0) {
    p[v][0] = par;
    for (int i = 1;; i++) {
      if (!p[p[v][i - 1]][i - 1]) break;
      p[v][i] = p[p[v][i - 1]][i - 1];
    }
  }
  for (int i = 0; i < g[v].size(); i++) {
    dfs(g[v][i], v, h + 1);
  }
  tout[v] = cur++;
}
int main() {
  cin >> n;
  vector<int> v;
  for (int i = 1; i < n + 1; i++) {
    int a;
    cin >> a;
    if (a == 0) {
      v.push_back(i);
      continue;
    }
    g[a].push_back(i);
  }
  for (int i = 0; i < v.size(); i++) {
    dfs(v[i], -1);
  }
  int m;
  cin >> m;
  vector<int> ans;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    int v = a;
    bool flag = false;
    for (int i = 0; i < 18; i++) {
      if ((1 << i) & b) v = p[v][i];
      if (v == 0) {
        flag = true;
        break;
      }
    }
    if (flag) {
      ans.push_back(0);
      continue;
    }
    int Len = H[a];
    int l = num[a], r = R[Len].size();
    while (r - l > 1) {
      int mid = (r + l) / 2;
      if (tin[R[Len][mid]] > tout[v])
        r = mid;
      else
        l = mid;
    }
    int end = l;
    l = -1, r = num[a];
    while (r - l > 1) {
      int mid = (r + l) / 2;
      if (tout[R[Len][mid]] < tin[v])
        l = mid;
      else
        r = mid;
    }
    int start = r;
    ans.push_back(end - start);
  }
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
}

### Prompt

Please formulate 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;
vector<vector<int>> g(1e5 + 1);
int n;
vector<int> H(1e5 + 1);
vector<vector<int>> p(1e5 + 1, vector<int>(18, 0));
vector<vector<int>> R(1e5 + 1);
vector<int> tin(1e5 + 1);
vector<int> tout(1e5 + 1);
vector<int> num(1e5 + 1);
int cur = 0;
void dfs(int v, int par, int h = 0) {
  H[v] = h;
  R[h].push_back(v);
  num[v] = R[h].size() - 1;
  tin[v] = cur++;
  if (h > 0) {
    p[v][0] = par;
    for (int i = 1;; i++) {
      if (!p[p[v][i - 1]][i - 1]) break;
      p[v][i] = p[p[v][i - 1]][i - 1];
    }
  }
  for (int i = 0; i < g[v].size(); i++) {
    dfs(g[v][i], v, h + 1);
  }
  tout[v] = cur++;
}
int main() {
  cin >> n;
  vector<int> v;
  for (int i = 1; i < n + 1; i++) {
    int a;
    cin >> a;
    if (a == 0) {
      v.push_back(i);
      continue;
    }
    g[a].push_back(i);
  }
  for (int i = 0; i < v.size(); i++) {
    dfs(v[i], -1);
  }
  int m;
  cin >> m;
  vector<int> ans;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    int v = a;
    bool flag = false;
    for (int i = 0; i < 18; i++) {
      if ((1 << i) & b) v = p[v][i];
      if (v == 0) {
        flag = true;
        break;
      }
    }
    if (flag) {
      ans.push_back(0);
      continue;
    }
    int Len = H[a];
    int l = num[a], r = R[Len].size();
    while (r - l > 1) {
      int mid = (r + l) / 2;
      if (tin[R[Len][mid]] > tout[v])
        r = mid;
      else
        l = mid;
    }
    int end = l;
    l = -1, r = num[a];
    while (r - l > 1) {
      int mid = (r + l) / 2;
      if (tout[R[Len][mid]] < tin[v])
        l = mid;
      else
        r = mid;
    }
    int start = r;
    ans.push_back(end - start);
  }
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
}

```