#include <bits/stdc++.h>
using namespace std;
template <class t>
t lcm(t a, t b) {
  return ((a * b) / __gcd(a, b));
}
long long n;
const long long logN = 20;
const long long maxn = 2e5;
long long parent[maxn][logN];
long long dfsnum = 0;
vector<long long> adjlist[maxn];
vector<long long> roots;
vector<long long> vertex;
vector<long long> oh, os, oe;
void digitize(long long arr[], long long val) {
  for (long long i = 0; i < logN; i++) {
    arr[i] = val % 2;
    val /= 2;
  }
}
long long height[maxn], st[maxn], en[maxn];
bool comp(long long a, long long b) {
  if (height[a] != height[b]) {
    return height[a] < height[b];
  } else {
    return st[a] < st[b];
  }
}
void input() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> parent[i][0];
    adjlist[parent[i][0]].push_back(i);
    if (parent[i][0] == 0) roots.push_back(i);
  }
}
void sparsetable() {
  for (long long j = 1; j < logN; j++) {
    for (long long i = 1; i <= n; i++) {
      parent[i][j] = parent[parent[i][j - 1]][j - 1];
    }
  }
}
long long calc(long long val, long long k) {
  long long hfind = height[val] + k;
  long long stpos, enpos;
  stpos = lower_bound(oh.begin(), oh.end(), hfind) - oh.begin();
  enpos = upper_bound(oh.begin(), oh.end(), hfind) - oh.begin();
  if (stpos == oh.size()) {
    return 0LL;
  }
  long long spos, epos;
  vector<long long>::iterator sit = os.begin();
  vector<long long>::iterator eit = os.begin();
  sit += stpos;
  eit += enpos;
  spos = lower_bound(sit, eit, st[val]) - os.begin();
  sit = oe.begin();
  eit = oe.begin();
  sit += stpos;
  eit += enpos;
  epos = upper_bound(sit, eit, en[val]) - oe.begin();
  if (spos >= epos)
    return 0LL;
  else
    return epos - spos - 1;
}
void solve() {
  long long u, v;
  cin >> u >> v;
  long long arr[logN];
  digitize(arr, v);
  for (long long i = 0; i < logN; i++) {
    if (arr[i] != 0) u = parent[u][i];
  }
  if (u == 0) {
    cout << 0;
    return;
  }
  cout << calc(u, v);
}
void dfs(long long cur) {
  st[cur] = dfsnum++;
  for (long long i = 0; i < adjlist[cur].size(); i++) {
    long long val = adjlist[cur][i];
    height[val] = height[cur] + 1;
    dfs(val);
  }
  en[cur] = dfsnum++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  input();
  sparsetable();
  for (long long i = 0; i < roots.size(); i++) dfs(roots[i]);
  for (long long i = 1; i <= n; i++) vertex.push_back(i);
  sort(vertex.begin(), vertex.end(), comp);
  for (long long i = 0; i < vertex.size(); i++) {
    long long cur = vertex[i];
    oh.push_back(height[cur]);
    os.push_back(st[cur]);
    oe.push_back(en[cur]);
  }
  long long q;
  cin >> q;
  while (q--) {
    solve();
    cout << " ";
  }
  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;
template <class t>
t lcm(t a, t b) {
  return ((a * b) / __gcd(a, b));
}
long long n;
const long long logN = 20;
const long long maxn = 2e5;
long long parent[maxn][logN];
long long dfsnum = 0;
vector<long long> adjlist[maxn];
vector<long long> roots;
vector<long long> vertex;
vector<long long> oh, os, oe;
void digitize(long long arr[], long long val) {
  for (long long i = 0; i < logN; i++) {
    arr[i] = val % 2;
    val /= 2;
  }
}
long long height[maxn], st[maxn], en[maxn];
bool comp(long long a, long long b) {
  if (height[a] != height[b]) {
    return height[a] < height[b];
  } else {
    return st[a] < st[b];
  }
}
void input() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> parent[i][0];
    adjlist[parent[i][0]].push_back(i);
    if (parent[i][0] == 0) roots.push_back(i);
  }
}
void sparsetable() {
  for (long long j = 1; j < logN; j++) {
    for (long long i = 1; i <= n; i++) {
      parent[i][j] = parent[parent[i][j - 1]][j - 1];
    }
  }
}
long long calc(long long val, long long k) {
  long long hfind = height[val] + k;
  long long stpos, enpos;
  stpos = lower_bound(oh.begin(), oh.end(), hfind) - oh.begin();
  enpos = upper_bound(oh.begin(), oh.end(), hfind) - oh.begin();
  if (stpos == oh.size()) {
    return 0LL;
  }
  long long spos, epos;
  vector<long long>::iterator sit = os.begin();
  vector<long long>::iterator eit = os.begin();
  sit += stpos;
  eit += enpos;
  spos = lower_bound(sit, eit, st[val]) - os.begin();
  sit = oe.begin();
  eit = oe.begin();
  sit += stpos;
  eit += enpos;
  epos = upper_bound(sit, eit, en[val]) - oe.begin();
  if (spos >= epos)
    return 0LL;
  else
    return epos - spos - 1;
}
void solve() {
  long long u, v;
  cin >> u >> v;
  long long arr[logN];
  digitize(arr, v);
  for (long long i = 0; i < logN; i++) {
    if (arr[i] != 0) u = parent[u][i];
  }
  if (u == 0) {
    cout << 0;
    return;
  }
  cout << calc(u, v);
}
void dfs(long long cur) {
  st[cur] = dfsnum++;
  for (long long i = 0; i < adjlist[cur].size(); i++) {
    long long val = adjlist[cur][i];
    height[val] = height[cur] + 1;
    dfs(val);
  }
  en[cur] = dfsnum++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  input();
  sparsetable();
  for (long long i = 0; i < roots.size(); i++) dfs(roots[i]);
  for (long long i = 1; i <= n; i++) vertex.push_back(i);
  sort(vertex.begin(), vertex.end(), comp);
  for (long long i = 0; i < vertex.size(); i++) {
    long long cur = vertex[i];
    oh.push_back(height[cur]);
    os.push_back(st[cur]);
    oe.push_back(en[cur]);
  }
  long long q;
  cin >> q;
  while (q--) {
    solve();
    cout << " ";
  }
  return 0;
}

```