#include <bits/stdc++.h>
using namespace std;
int pare[100000], dfsn[100000], P[100000][20], n, m, level[100000], ND[100000],
    dfst;
vector<vector<int> > height, graf;
void dfs(int u) {
  dfsn[u] = dfst++;
  height[level[u]].push_back(dfsn[u]);
  ND[u] = 1;
  for (int i = 0; i < graf[u].size(); ++i) {
    int v = graf[u][i];
    if (dfsn[v] == -1) {
      level[v] = level[u] + 1;
      dfs(v);
      ND[u] += ND[v];
    }
  }
}
int anc(int v, int p) {
  for (int k = 19; k >= 0; --k)
    if (1 << k <= p) {
      p -= 1 << k;
      if (v == -1) return -1;
      v = P[v][k];
    }
  return v;
}
int main() {
  cin >> n;
  dfst = 0;
  graf = height = vector<vector<int> >(n);
  for (int i = 0; i < n; ++i) {
    cin >> pare[i];
    pare[i]--;
    dfsn[i] = -1;
    if (pare[i] != -1) graf[pare[i]].push_back(i);
  }
  for (int i = 0; i < n; ++i)
    if (dfsn[i] == -1) dfs(i);
  for (int i = 0; i < n; ++i) P[i][0] = pare[i];
  for (int j = 1; 1 << j < n; ++j)
    for (int i = 0; i < n; ++i)
      P[i][j] = (P[i][j - 1] == -1) ? -1 : P[P[i][j - 1]][j - 1];
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int v, p;
    cin >> v >> p;
    v--;
    int u = anc(v, p);
    if (u != -1)
      cout << upper_bound(height[level[v]].begin(), height[level[v]].end(),
                          dfsn[u] + ND[u] - 1) -
                  lower_bound(height[level[v]].begin(), height[level[v]].end(),
                              dfsn[u]) -
                  1
           << endl;
    else
      cout << 0 << endl;
  }
}

### 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;
int pare[100000], dfsn[100000], P[100000][20], n, m, level[100000], ND[100000],
    dfst;
vector<vector<int> > height, graf;
void dfs(int u) {
  dfsn[u] = dfst++;
  height[level[u]].push_back(dfsn[u]);
  ND[u] = 1;
  for (int i = 0; i < graf[u].size(); ++i) {
    int v = graf[u][i];
    if (dfsn[v] == -1) {
      level[v] = level[u] + 1;
      dfs(v);
      ND[u] += ND[v];
    }
  }
}
int anc(int v, int p) {
  for (int k = 19; k >= 0; --k)
    if (1 << k <= p) {
      p -= 1 << k;
      if (v == -1) return -1;
      v = P[v][k];
    }
  return v;
}
int main() {
  cin >> n;
  dfst = 0;
  graf = height = vector<vector<int> >(n);
  for (int i = 0; i < n; ++i) {
    cin >> pare[i];
    pare[i]--;
    dfsn[i] = -1;
    if (pare[i] != -1) graf[pare[i]].push_back(i);
  }
  for (int i = 0; i < n; ++i)
    if (dfsn[i] == -1) dfs(i);
  for (int i = 0; i < n; ++i) P[i][0] = pare[i];
  for (int j = 1; 1 << j < n; ++j)
    for (int i = 0; i < n; ++i)
      P[i][j] = (P[i][j - 1] == -1) ? -1 : P[P[i][j - 1]][j - 1];
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int v, p;
    cin >> v >> p;
    v--;
    int u = anc(v, p);
    if (u != -1)
      cout << upper_bound(height[level[v]].begin(), height[level[v]].end(),
                          dfsn[u] + ND[u] - 1) -
                  lower_bound(height[level[v]].begin(), height[level[v]].end(),
                              dfsn[u]) -
                  1
           << endl;
    else
      cout << 0 << endl;
  }
}

```