#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 9, lg = 20;
vector<int> g[MAX], h[MAX];
int cnt, d[MAX], s[MAX], f[MAX], par[MAX][lg];
void dfs(int v) {
  s[v] = cnt++;
  h[d[v]].push_back(s[v]);
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    d[u] = d[v] + 1;
    par[u][0] = v;
    for (int i = 1; i < lg; i++) par[u][i] = par[par[u][i - 1]][i - 1];
    dfs(u);
  }
  f[v] = cnt;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n, v, q, p;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v, g[v].push_back(i + 1);
  dfs(0);
  cin >> q;
  while (q--) {
    cin >> v >> p;
    for (int i = 0; i < lg; i++)
      if (p & (1 << i)) v = par[v][i];
    if (!v) {
      cout << 0 << ' ';
      continue;
    }
    cout << (upper_bound(h[d[v] + p].begin(), h[d[v] + p].end(), f[v]) -
             h[d[v] + p].begin()) -
                (upper_bound(h[d[v] + p].begin(), h[d[v] + p].end(), s[v]) -
                 h[d[v] + p].begin()) -
                1
         << ' ';
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MAX = 1e5 + 9, lg = 20;
vector<int> g[MAX], h[MAX];
int cnt, d[MAX], s[MAX], f[MAX], par[MAX][lg];
void dfs(int v) {
  s[v] = cnt++;
  h[d[v]].push_back(s[v]);
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    d[u] = d[v] + 1;
    par[u][0] = v;
    for (int i = 1; i < lg; i++) par[u][i] = par[par[u][i - 1]][i - 1];
    dfs(u);
  }
  f[v] = cnt;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n, v, q, p;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> v, g[v].push_back(i + 1);
  dfs(0);
  cin >> q;
  while (q--) {
    cin >> v >> p;
    for (int i = 0; i < lg; i++)
      if (p & (1 << i)) v = par[v][i];
    if (!v) {
      cout << 0 << ' ';
      continue;
    }
    cout << (upper_bound(h[d[v] + p].begin(), h[d[v] + p].end(), f[v]) -
             h[d[v] + p].begin()) -
                (upper_bound(h[d[v] + p].begin(), h[d[v] + p].end(), s[v]) -
                 h[d[v] + p].begin()) -
                1
         << ' ';
  }
}

```