#include <bits/stdc++.h>
using namespace std;
int n, q, timer;
int tin[100001], tout[100001];
vector<int> adj[100001];
int par[100001][20];
int level[100001];
vector<int> rlvl[100001];
int vis[100001];
void dfs(int v, int p, int d) {
  par[v][0] = p;
  vis[v] = 1;
  tin[v] = ++timer;
  level[v] = d + 1;
  rlvl[d + 1].push_back(timer);
  for (int i : adj[v]) {
    if (i != p) {
      dfs(i, v, d + 1);
    }
  }
  tout[v] = ++timer;
}
void preprocess() {
  for (int i = 1; i <= 18; i++) {
    for (int nd = 1; nd <= n; nd++) {
      if (par[nd][i - 1] != -1) {
        par[nd][i] = par[par[nd][i - 1]][i - 1];
      }
    }
  }
}
int kth(int k, int v) {
  for (int i = 0; i < 18; i++) {
    if (k & (1 << i)) {
      v = par[v][i];
      if (v == -1) {
        break;
      }
    }
  }
  return v;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    adj[x].push_back(i);
  }
  vis[0] = 1;
  memset(par, -1, sizeof(par));
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() > 0 && !vis[i]) {
      dfs(i, 0, 0);
    }
  }
  preprocess();
  int q;
  cin >> q;
  vector<int> res;
  while (q--) {
    int v, p;
    cin >> v >> p;
    int u = kth(p, v);
    if (u == -1 || u == 0) {
      res.push_back(0);
      continue;
    }
    int dep = level[v];
    int grt = rlvl[dep].end() -
              lower_bound(rlvl[dep].begin(), rlvl[dep].end(), tout[u]);
    int sml = rlvl[dep].end() -
              lower_bound(rlvl[dep].begin(), rlvl[dep].end(), tin[u]);
    res.push_back(sml - grt - 1);
  }
  for (int j : res) cout << j << " ";
}

### 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, q, timer;
int tin[100001], tout[100001];
vector<int> adj[100001];
int par[100001][20];
int level[100001];
vector<int> rlvl[100001];
int vis[100001];
void dfs(int v, int p, int d) {
  par[v][0] = p;
  vis[v] = 1;
  tin[v] = ++timer;
  level[v] = d + 1;
  rlvl[d + 1].push_back(timer);
  for (int i : adj[v]) {
    if (i != p) {
      dfs(i, v, d + 1);
    }
  }
  tout[v] = ++timer;
}
void preprocess() {
  for (int i = 1; i <= 18; i++) {
    for (int nd = 1; nd <= n; nd++) {
      if (par[nd][i - 1] != -1) {
        par[nd][i] = par[par[nd][i - 1]][i - 1];
      }
    }
  }
}
int kth(int k, int v) {
  for (int i = 0; i < 18; i++) {
    if (k & (1 << i)) {
      v = par[v][i];
      if (v == -1) {
        break;
      }
    }
  }
  return v;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    adj[x].push_back(i);
  }
  vis[0] = 1;
  memset(par, -1, sizeof(par));
  for (int i = 1; i <= n; i++) {
    if (adj[i].size() > 0 && !vis[i]) {
      dfs(i, 0, 0);
    }
  }
  preprocess();
  int q;
  cin >> q;
  vector<int> res;
  while (q--) {
    int v, p;
    cin >> v >> p;
    int u = kth(p, v);
    if (u == -1 || u == 0) {
      res.push_back(0);
      continue;
    }
    int dep = level[v];
    int grt = rlvl[dep].end() -
              lower_bound(rlvl[dep].begin(), rlvl[dep].end(), tout[u]);
    int sml = rlvl[dep].end() -
              lower_bound(rlvl[dep].begin(), rlvl[dep].end(), tin[u]);
    res.push_back(sml - grt - 1);
  }
  for (int j : res) cout << j << " ";
}

```