#include <bits/stdc++.h>
using namespace std;
const int M = 100000 + 10;
struct EDGE {
  int u, v;
} edge[M];
vector<EDGE> adj[M];
vector<int> root;
int timestamp = 0;
int timestamparray[M];
int revtimestamparray[M];
int lefttimeborder[M];
int righttimeborder[M];
int depth[M];
vector<int> samedepth[M];
void dfs(int cur, int par) {
  timestamp++;
  timestamparray[cur] = timestamp;
  revtimestamparray[timestamp] = cur;
  lefttimeborder[cur] = timestamp + 1;
  depth[cur] = depth[par] + 1;
  samedepth[depth[cur]].push_back(timestamp);
  for (vector<EDGE>::iterator it = adj[cur].begin(); it != adj[cur].end();
       it++) {
    int u = (*it).u;
    int v = (*it).v;
    int nxt = (u == cur ? v : u);
    if (nxt != par) {
      dfs(nxt, cur);
    }
  }
  righttimeborder[cur] = timestamp;
}
int main() {
  int n;
  cin >> n;
  int nume = 0;
  for (int i = 1; i <= n; i++) {
    int u;
    cin >> u;
    if (u != 0) {
      edge[nume].u = u;
      edge[nume].v = i;
      adj[u].push_back(edge[nume]);
      adj[i].push_back(edge[nume]);
      nume++;
    } else {
      root.push_back(i);
    }
  }
  for (vector<int>::iterator it = root.begin(); it != root.end(); it++) {
    int cur = (*it);
    dfs(cur, 0);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, d;
    cin >> v >> d;
    if (depth[v] - 1 < d) {
      cout << "0 ";
    } else {
      int targetdepth = depth[v] - d;
      int tid = lower_bound(samedepth[targetdepth].begin(),
                            samedepth[targetdepth].end(), timestamparray[v]) -
                samedepth[targetdepth].begin();
      tid--;
      int ttime = samedepth[targetdepth][tid];
      int parnode = revtimestamparray[ttime];
      int ans = 0;
      int lefttime = lefttimeborder[parnode];
      int righttime = righttimeborder[parnode];
      int leftid = lower_bound(samedepth[depth[v]].begin(),
                               samedepth[depth[v]].end(), lefttime) -
                   samedepth[depth[v]].begin();
      int rightid = upper_bound(samedepth[depth[v]].begin(),
                                samedepth[depth[v]].end(), righttime) -
                    samedepth[depth[v]].begin() - 1;
      ans = rightid - leftid + 1 - 1;
      cout << ans << " ";
    }
  }
  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 M = 100000 + 10;
struct EDGE {
  int u, v;
} edge[M];
vector<EDGE> adj[M];
vector<int> root;
int timestamp = 0;
int timestamparray[M];
int revtimestamparray[M];
int lefttimeborder[M];
int righttimeborder[M];
int depth[M];
vector<int> samedepth[M];
void dfs(int cur, int par) {
  timestamp++;
  timestamparray[cur] = timestamp;
  revtimestamparray[timestamp] = cur;
  lefttimeborder[cur] = timestamp + 1;
  depth[cur] = depth[par] + 1;
  samedepth[depth[cur]].push_back(timestamp);
  for (vector<EDGE>::iterator it = adj[cur].begin(); it != adj[cur].end();
       it++) {
    int u = (*it).u;
    int v = (*it).v;
    int nxt = (u == cur ? v : u);
    if (nxt != par) {
      dfs(nxt, cur);
    }
  }
  righttimeborder[cur] = timestamp;
}
int main() {
  int n;
  cin >> n;
  int nume = 0;
  for (int i = 1; i <= n; i++) {
    int u;
    cin >> u;
    if (u != 0) {
      edge[nume].u = u;
      edge[nume].v = i;
      adj[u].push_back(edge[nume]);
      adj[i].push_back(edge[nume]);
      nume++;
    } else {
      root.push_back(i);
    }
  }
  for (vector<int>::iterator it = root.begin(); it != root.end(); it++) {
    int cur = (*it);
    dfs(cur, 0);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, d;
    cin >> v >> d;
    if (depth[v] - 1 < d) {
      cout << "0 ";
    } else {
      int targetdepth = depth[v] - d;
      int tid = lower_bound(samedepth[targetdepth].begin(),
                            samedepth[targetdepth].end(), timestamparray[v]) -
                samedepth[targetdepth].begin();
      tid--;
      int ttime = samedepth[targetdepth][tid];
      int parnode = revtimestamparray[ttime];
      int ans = 0;
      int lefttime = lefttimeborder[parnode];
      int righttime = righttimeborder[parnode];
      int leftid = lower_bound(samedepth[depth[v]].begin(),
                               samedepth[depth[v]].end(), lefttime) -
                   samedepth[depth[v]].begin();
      int rightid = upper_bound(samedepth[depth[v]].begin(),
                                samedepth[depth[v]].end(), righttime) -
                    samedepth[depth[v]].begin() - 1;
      ans = rightid - leftid + 1 - 1;
      cout << ans << " ";
    }
  }
  cout << endl;
  return 0;
}

```