#include <bits/stdc++.h>
using namespace std;
map<pair<long long, long long>, long long> mp;
const int N = +1e+5 + 5;
const int LG = 20;
vector<long long> g[N];
long long parent[LG][N];
long long timer = 0;
long long tin[N], tout[N], level[N];
vector<long long> vv[N];
long long n, p;
long long cnt = 0;
long long arr[N];
void dfs(long long v, long long par, long long lvl) {
  arr[v] = cnt;
  tin[v] = ++timer;
  parent[0][v] = par;
  level[v] = lvl;
  vv[lvl].push_back(timer);
  for (auto &u : g[v]) {
    if (u == par) continue;
    dfs(u, v, lvl + 1);
  }
  tout[v] = timer;
}
bool is_ancestor(int u, int v) {
  return (tin[u] <= tin[v] and tout[u] >= tout[v]);
}
void precompute() {
  for (int i = 1; i < LG; i++) {
    for (int j = 1; j <= n; j++) {
      if (parent[i - 1][j]) {
        parent[i][j] = parent[i - 1][parent[i - 1][j]];
      }
    }
  }
}
long long kthancestor(long long v, long long p) {
  long long curr = v;
  for (long long i = 20; i >= 0; i--) {
    if (1 << i & p) curr = parent[i][curr];
  }
  return curr;
}
void solve() {
  long long x;
  cin >> n;
  vector<long long> v;
  for (long long i = 1; i <= n; i++) {
    cin >> x;
    if (x == 0) {
      v.push_back(i);
    } else {
      g[x].push_back(i);
      g[i].push_back(x);
    }
  }
  for (long long x : v) {
    dfs(x, 0, 0);
    cnt++;
  }
  precompute();
  long long m;
  cin >> m;
  for (long long i = 1; i <= m; i++) {
    long long v, p;
    cin >> v >> p;
    long long x = kthancestor(v, p);
    if (x == 0) {
      cout << 0 << " ";
    } else {
      long long y = level[v];
      long long ans = upper_bound(vv[y].begin(), vv[y].end(), tout[x]) -
                      lower_bound(vv[y].begin(), vv[y].end(), tin[x]);
      ans--;
      cout << ans << " ";
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
map<pair<long long, long long>, long long> mp;
const int N = +1e+5 + 5;
const int LG = 20;
vector<long long> g[N];
long long parent[LG][N];
long long timer = 0;
long long tin[N], tout[N], level[N];
vector<long long> vv[N];
long long n, p;
long long cnt = 0;
long long arr[N];
void dfs(long long v, long long par, long long lvl) {
  arr[v] = cnt;
  tin[v] = ++timer;
  parent[0][v] = par;
  level[v] = lvl;
  vv[lvl].push_back(timer);
  for (auto &u : g[v]) {
    if (u == par) continue;
    dfs(u, v, lvl + 1);
  }
  tout[v] = timer;
}
bool is_ancestor(int u, int v) {
  return (tin[u] <= tin[v] and tout[u] >= tout[v]);
}
void precompute() {
  for (int i = 1; i < LG; i++) {
    for (int j = 1; j <= n; j++) {
      if (parent[i - 1][j]) {
        parent[i][j] = parent[i - 1][parent[i - 1][j]];
      }
    }
  }
}
long long kthancestor(long long v, long long p) {
  long long curr = v;
  for (long long i = 20; i >= 0; i--) {
    if (1 << i & p) curr = parent[i][curr];
  }
  return curr;
}
void solve() {
  long long x;
  cin >> n;
  vector<long long> v;
  for (long long i = 1; i <= n; i++) {
    cin >> x;
    if (x == 0) {
      v.push_back(i);
    } else {
      g[x].push_back(i);
      g[i].push_back(x);
    }
  }
  for (long long x : v) {
    dfs(x, 0, 0);
    cnt++;
  }
  precompute();
  long long m;
  cin >> m;
  for (long long i = 1; i <= m; i++) {
    long long v, p;
    cin >> v >> p;
    long long x = kthancestor(v, p);
    if (x == 0) {
      cout << 0 << " ";
    } else {
      long long y = level[v];
      long long ans = upper_bound(vv[y].begin(), vv[y].end(), tout[x]) -
                      lower_bound(vv[y].begin(), vv[y].end(), tin[x]);
      ans--;
      cout << ans << " ";
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  t = 1;
  while (t--) solve();
  return 0;
}

```