#include <bits/stdc++.h>
using namespace std;
vector<int> node[100001];
int timestamp;
int height[100001];
int dfsorder[110001];
int timemap[110001];
int dfsleft[100001];
int dfsright[100001];
vector<int> depthorder[100001];
void dfs(int now, int h) {
  height[now] = h;
  dfsorder[now] = timestamp;
  timemap[timestamp] = now;
  dfsleft[now] = timestamp + 1;
  depthorder[h].push_back(timestamp);
  timestamp++;
  for (int i = 0; i < node[now].size(); i++) {
    dfs(node[now][i], h + 1);
  }
  dfsright[now] = timestamp - 1;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  memset(height, -1, sizeof(height));
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int u;
    cin >> u;
    if (u != 0) {
      node[u].push_back(i);
    } else
      height[i] = 0;
  }
  timestamp = 0;
  for (int i = 1; i <= n; i++) {
    if (height[i] == 0) {
      dfs(i, 0);
    }
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, p;
    cin >> v >> p;
    int paheight = height[v] - p;
    if (paheight < 0) {
      cout << 0 << " ";
      continue;
    }
    int pa = timemap[*(upper_bound(depthorder[paheight].begin(),
                                   depthorder[paheight].end(), dfsorder[v]) -
                       1)];
    int l = dfsleft[pa], r = dfsright[pa];
    int ans = upper_bound(depthorder[height[v]].begin(),
                          depthorder[height[v]].end(), r) -
              lower_bound(depthorder[height[v]].begin(),
                          depthorder[height[v]].end(), l) -
              1;
    cout << ans << " ";
  }
  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;
vector<int> node[100001];
int timestamp;
int height[100001];
int dfsorder[110001];
int timemap[110001];
int dfsleft[100001];
int dfsright[100001];
vector<int> depthorder[100001];
void dfs(int now, int h) {
  height[now] = h;
  dfsorder[now] = timestamp;
  timemap[timestamp] = now;
  dfsleft[now] = timestamp + 1;
  depthorder[h].push_back(timestamp);
  timestamp++;
  for (int i = 0; i < node[now].size(); i++) {
    dfs(node[now][i], h + 1);
  }
  dfsright[now] = timestamp - 1;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  memset(height, -1, sizeof(height));
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int u;
    cin >> u;
    if (u != 0) {
      node[u].push_back(i);
    } else
      height[i] = 0;
  }
  timestamp = 0;
  for (int i = 1; i <= n; i++) {
    if (height[i] == 0) {
      dfs(i, 0);
    }
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, p;
    cin >> v >> p;
    int paheight = height[v] - p;
    if (paheight < 0) {
      cout << 0 << " ";
      continue;
    }
    int pa = timemap[*(upper_bound(depthorder[paheight].begin(),
                                   depthorder[paheight].end(), dfsorder[v]) -
                       1)];
    int l = dfsleft[pa], r = dfsright[pa];
    int ans = upper_bound(depthorder[height[v]].begin(),
                          depthorder[height[v]].end(), r) -
              lower_bound(depthorder[height[v]].begin(),
                          depthorder[height[v]].end(), l) -
              1;
    cout << ans << " ";
  }
  return 0;
}

```