#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int M = 17;
int n, q;
int par[N];
vector<int> adl[N];
int upar[N], sz[N];
int dep[N];
vector<int> tree[N];
int tin[N], tout[N];
vector<int> euler[N];
vector<unordered_map<int, int>> t[N];
int up[N][M];
bool is_ancestor(int a, int b) {
  return tin[a] <= tin[b] && tout[b] <= tout[a];
}
int find(int a) {
  if (upar[a] == a) return a;
  return upar[a] = find(upar[a]);
}
void make(int a, int b) {
  a = find(a);
  b = find(b);
  if (a == b) return;
  sz[a] += sz[b];
  upar[b] = a;
}
int timer[N];
void dfs_tur(int node, int p) {
  dep[node] = dep[p] + 1;
  tin[node] = ++timer[find(node)];
  euler[find(node)].push_back(dep[node]);
  up[node][0] = p;
  for (int i = 1; i < M; i++) {
    up[node][i] = up[up[node][i - 1]][i - 1];
  }
  for (auto x : adl[node]) {
    if (x == p) continue;
    dfs_tur(x, node);
  }
  tout[node] = timer[find(node)];
}
int uni;
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[uni][v][euler[uni][tl]]++;
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm);
  build(v * 2 + 1, tm + 1, tr);
  for (auto x : t[uni][v * 2]) {
    t[uni][v][x.first] = x.second;
  }
  for (auto x : t[uni][v * 2 + 1]) {
    t[uni][v][x.first] += x.second;
  }
}
int val;
int ask(int v, int tl, int tr, int l, int r) {
  if (l > r)
    return 0;
  else if (l <= tl && tr <= r) {
    return t[uni][v][val];
  }
  int tm = (tl + tr) / 2;
  return ask(v * 2, tl, tm, l, min(tm, r)) +
         ask(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
int xth_ancestor(int node, int x) {
  int cur = 0;
  for (int i = M - 1; i >= 0; i--) {
    int plus = (1 << i);
    if (cur + plus <= x) {
      node = up[node][i];
      cur += plus;
    }
  }
  return node;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    sz[i] = 1;
    upar[i] = i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> par[i];
    if (par[i] != 0) {
      adl[par[i]].push_back(i);
      adl[i].push_back(par[i]);
      make(par[i], i);
    }
  }
  for (int i = 1; i <= n; i++) {
    tree[find(i)].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (upar[i] == i) {
      euler[i].push_back(-1);
      dfs_tur(i, i);
      t[i].resize(timer[i] * 4 + 10);
      uni = i;
      build(1, 1, timer[i]);
    }
  }
  cin >> q;
  while (q--) {
    int v, p;
    cin >> v >> p;
    if (dep[v] <= p) {
      cout << 0 << " ";
      continue;
    }
    v = xth_ancestor(v, p);
    uni = find(v);
    val = dep[v] + p;
    cout << ask(1, 1, timer[uni], tin[v], tout[v]) - 1 << " ";
  }
  cout << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 1e5 + 10;
const int M = 17;
int n, q;
int par[N];
vector<int> adl[N];
int upar[N], sz[N];
int dep[N];
vector<int> tree[N];
int tin[N], tout[N];
vector<int> euler[N];
vector<unordered_map<int, int>> t[N];
int up[N][M];
bool is_ancestor(int a, int b) {
  return tin[a] <= tin[b] && tout[b] <= tout[a];
}
int find(int a) {
  if (upar[a] == a) return a;
  return upar[a] = find(upar[a]);
}
void make(int a, int b) {
  a = find(a);
  b = find(b);
  if (a == b) return;
  sz[a] += sz[b];
  upar[b] = a;
}
int timer[N];
void dfs_tur(int node, int p) {
  dep[node] = dep[p] + 1;
  tin[node] = ++timer[find(node)];
  euler[find(node)].push_back(dep[node]);
  up[node][0] = p;
  for (int i = 1; i < M; i++) {
    up[node][i] = up[up[node][i - 1]][i - 1];
  }
  for (auto x : adl[node]) {
    if (x == p) continue;
    dfs_tur(x, node);
  }
  tout[node] = timer[find(node)];
}
int uni;
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[uni][v][euler[uni][tl]]++;
    return;
  }
  int tm = (tl + tr) / 2;
  build(v * 2, tl, tm);
  build(v * 2 + 1, tm + 1, tr);
  for (auto x : t[uni][v * 2]) {
    t[uni][v][x.first] = x.second;
  }
  for (auto x : t[uni][v * 2 + 1]) {
    t[uni][v][x.first] += x.second;
  }
}
int val;
int ask(int v, int tl, int tr, int l, int r) {
  if (l > r)
    return 0;
  else if (l <= tl && tr <= r) {
    return t[uni][v][val];
  }
  int tm = (tl + tr) / 2;
  return ask(v * 2, tl, tm, l, min(tm, r)) +
         ask(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
}
int xth_ancestor(int node, int x) {
  int cur = 0;
  for (int i = M - 1; i >= 0; i--) {
    int plus = (1 << i);
    if (cur + plus <= x) {
      node = up[node][i];
      cur += plus;
    }
  }
  return node;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    sz[i] = 1;
    upar[i] = i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> par[i];
    if (par[i] != 0) {
      adl[par[i]].push_back(i);
      adl[i].push_back(par[i]);
      make(par[i], i);
    }
  }
  for (int i = 1; i <= n; i++) {
    tree[find(i)].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (upar[i] == i) {
      euler[i].push_back(-1);
      dfs_tur(i, i);
      t[i].resize(timer[i] * 4 + 10);
      uni = i;
      build(1, 1, timer[i]);
    }
  }
  cin >> q;
  while (q--) {
    int v, p;
    cin >> v >> p;
    if (dep[v] <= p) {
      cout << 0 << " ";
      continue;
    }
    v = xth_ancestor(v, p);
    uni = find(v);
    val = dep[v] + p;
    cout << ask(1, 1, timer[uni], tin[v], tout[v]) - 1 << " ";
  }
  cout << '\n';
  return 0;
}

```