#include <bits/stdc++.h>
using namespace std;
const int ms = 1e5 + 5, mlg = 23;
int par[ms][mlg + 1], lvl[ms], n;
int id = 1;
vector<int> g[ms];
vector<int> h[ms];
int in[ms], out[ms];
void dfs(int v, int p, int l = 0) {
  lvl[v] = l;
  in[v] = id++;
  h[l].push_back(id);
  par[v][0] = p;
  for (int k = 1; k <= mlg; ++k) {
    par[v][k] = par[par[v][k - 1]][k - 1];
  }
  for (int u : g[v]) {
    if (u != p) {
      dfs(u, v, l + 1);
    }
  }
  out[v] = id++;
}
int up(int a, int p) {
  for (int i = mlg; i >= 0; --i) {
    if (p & (1 << i)) {
      a = par[a][i];
    }
  }
  return a;
}
int main() {
  cin >> n;
  vector<int> root;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    if (x == 0) {
      root.push_back(i);
    } else {
      g[x - 1].push_back(i);
    }
  }
  for (int i = 0; i < root.size(); ++i) {
    dfs(root[i], root[i]);
  }
  int m;
  cin >> m;
  while (m--) {
    int v, p;
    scanf("%d%d", &v, &p);
    v--;
    if (lvl[v] < p) {
      cout << "0 ";
    } else {
      int parent = up(v, p);
      int ans = upper_bound(h[lvl[v]].begin(), h[lvl[v]].end(), out[parent]) -
                lower_bound(h[lvl[v]].begin(), h[lvl[v]].end(), in[parent]);
      ans--;
      cout << ans << ' ';
    }
  }
  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;
const int ms = 1e5 + 5, mlg = 23;
int par[ms][mlg + 1], lvl[ms], n;
int id = 1;
vector<int> g[ms];
vector<int> h[ms];
int in[ms], out[ms];
void dfs(int v, int p, int l = 0) {
  lvl[v] = l;
  in[v] = id++;
  h[l].push_back(id);
  par[v][0] = p;
  for (int k = 1; k <= mlg; ++k) {
    par[v][k] = par[par[v][k - 1]][k - 1];
  }
  for (int u : g[v]) {
    if (u != p) {
      dfs(u, v, l + 1);
    }
  }
  out[v] = id++;
}
int up(int a, int p) {
  for (int i = mlg; i >= 0; --i) {
    if (p & (1 << i)) {
      a = par[a][i];
    }
  }
  return a;
}
int main() {
  cin >> n;
  vector<int> root;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    if (x == 0) {
      root.push_back(i);
    } else {
      g[x - 1].push_back(i);
    }
  }
  for (int i = 0; i < root.size(); ++i) {
    dfs(root[i], root[i]);
  }
  int m;
  cin >> m;
  while (m--) {
    int v, p;
    scanf("%d%d", &v, &p);
    v--;
    if (lvl[v] < p) {
      cout << "0 ";
    } else {
      int parent = up(v, p);
      int ans = upper_bound(h[lvl[v]].begin(), h[lvl[v]].end(), out[parent]) -
                lower_bound(h[lvl[v]].begin(), h[lvl[v]].end(), in[parent]);
      ans--;
      cout << ans << ' ';
    }
  }
  return 0;
}

```