#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using i3 = pair<int, ii>;
using li = pair<ll, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vli = vector<li>;
using vii = vector<ii>;
using vi3 = vector<i3>;
const int N = 1e5 + 5, L = 20;
const ll INF = 1e17;
const double eps = 1e-12, PI = acos(-1);
int n, m;
int to[N][L];
vi start, adj[N], level[N];
int in[N], out[N], timer = 0;
int dep[N];
void dfs(int u, int p) {
  to[u][0] = p;
  dep[u] = 1 + dep[p];
  in[u] = ++timer;
  level[dep[u]].push_back(timer);
  for (int v : adj[u]) {
    dfs(v, u);
  }
  out[u] = timer;
}
void solve(int cs) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p) {
      adj[p].push_back(i);
    } else
      start.push_back(i);
  }
  int sz = start.size();
  for (int i = 0; i < sz; i++) {
    dfs(start[i], 0);
  }
  for (int j = 1; j < L; j++)
    for (int i = 1; i <= n; i++) to[i][j] = to[to[i][j - 1]][j - 1];
  int q;
  scanf("%d", &q);
  for (; q--;) {
    int k, v;
    scanf("%d %d", &v, &k);
    int dx = dep[v];
    for (int i = 0; i < L - 1; i++) {
      if (k & (1 << i)) v = to[v][i];
    }
    int L = lower_bound(level[dx].begin(), level[dx].end(), in[v]) -
            level[dx].begin();
    int R = upper_bound(level[dx].begin(), level[dx].end(), out[v]) -
            level[dx].begin();
    printf("%d ", R - L - (R != L));
  }
}
int main() {
  int t = 1;
  for (int cs = 1; t--; cs++) {
    solve(cs);
  }
  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;
using ll = long long;
using ii = pair<int, int>;
using i3 = pair<int, ii>;
using li = pair<ll, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vli = vector<li>;
using vii = vector<ii>;
using vi3 = vector<i3>;
const int N = 1e5 + 5, L = 20;
const ll INF = 1e17;
const double eps = 1e-12, PI = acos(-1);
int n, m;
int to[N][L];
vi start, adj[N], level[N];
int in[N], out[N], timer = 0;
int dep[N];
void dfs(int u, int p) {
  to[u][0] = p;
  dep[u] = 1 + dep[p];
  in[u] = ++timer;
  level[dep[u]].push_back(timer);
  for (int v : adj[u]) {
    dfs(v, u);
  }
  out[u] = timer;
}
void solve(int cs) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int p;
    scanf("%d", &p);
    if (p) {
      adj[p].push_back(i);
    } else
      start.push_back(i);
  }
  int sz = start.size();
  for (int i = 0; i < sz; i++) {
    dfs(start[i], 0);
  }
  for (int j = 1; j < L; j++)
    for (int i = 1; i <= n; i++) to[i][j] = to[to[i][j - 1]][j - 1];
  int q;
  scanf("%d", &q);
  for (; q--;) {
    int k, v;
    scanf("%d %d", &v, &k);
    int dx = dep[v];
    for (int i = 0; i < L - 1; i++) {
      if (k & (1 << i)) v = to[v][i];
    }
    int L = lower_bound(level[dx].begin(), level[dx].end(), in[v]) -
            level[dx].begin();
    int R = upper_bound(level[dx].begin(), level[dx].end(), out[v]) -
            level[dx].begin();
    printf("%d ", R - L - (R != L));
  }
}
int main() {
  int t = 1;
  for (int cs = 1; t--; cs++) {
    solve(cs);
  }
  return 0;
}

```