#include <bits/stdc++.h>
using namespace std;
const int depth = 20;
int pr[100500][depth];
vector<int> children[100500];
list<pair<int, int> > tr[100500];
int curr_len = 0;
int a[100500];
int v[100500];
int p[100500];
void dfs(int u) {
  curr_len++;
  for (int i = 1; i < depth; i++) pr[u][i] = pr[pr[u][i - 1]][i - 1];
  for (int i = 0; i < children[u].size(); i++) dfs(children[u][i]);
  curr_len--;
}
void dfs2(int u) {
  curr_len++;
  a[curr_len]++;
  for (auto it = tr[u].begin(); it != tr[u].end(); it++)
    it->second = -a[curr_len + it->first];
  for (int i = 0; i < children[u].size(); i++) dfs2(children[u][i]);
  for (auto it = tr[u].begin(); it != tr[u].end(); it++)
    it->second += a[curr_len + it->first];
  curr_len--;
}
int get_pred(int u, int len) {
  for (int i = 0; i < 20; i++) {
    if (len & 1) u = pr[u][i];
    len /= 2;
  }
  return u;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &pr[i + 1][0]);
    children[pr[i + 1][0]].push_back(i + 1);
  }
  dfs(0);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", v + i, p + i);
    v[i] = get_pred(v[i], p[i]);
    tr[v[i]].push_back(make_pair(p[i], 0));
  }
  dfs2(0);
  for (int i = 0; i < m; i++)
    if (v[i] == 0)
      printf("0 ");
    else {
      printf("%d ", tr[v[i]].front().second - 1);
      tr[v[i]].pop_front();
    }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 depth = 20;
int pr[100500][depth];
vector<int> children[100500];
list<pair<int, int> > tr[100500];
int curr_len = 0;
int a[100500];
int v[100500];
int p[100500];
void dfs(int u) {
  curr_len++;
  for (int i = 1; i < depth; i++) pr[u][i] = pr[pr[u][i - 1]][i - 1];
  for (int i = 0; i < children[u].size(); i++) dfs(children[u][i]);
  curr_len--;
}
void dfs2(int u) {
  curr_len++;
  a[curr_len]++;
  for (auto it = tr[u].begin(); it != tr[u].end(); it++)
    it->second = -a[curr_len + it->first];
  for (int i = 0; i < children[u].size(); i++) dfs2(children[u][i]);
  for (auto it = tr[u].begin(); it != tr[u].end(); it++)
    it->second += a[curr_len + it->first];
  curr_len--;
}
int get_pred(int u, int len) {
  for (int i = 0; i < 20; i++) {
    if (len & 1) u = pr[u][i];
    len /= 2;
  }
  return u;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &pr[i + 1][0]);
    children[pr[i + 1][0]].push_back(i + 1);
  }
  dfs(0);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", v + i, p + i);
    v[i] = get_pred(v[i], p[i]);
    tr[v[i]].push_back(make_pair(p[i], 0));
  }
  dfs2(0);
  for (int i = 0; i < m; i++)
    if (v[i] == 0)
      printf("0 ");
    else {
      printf("%d ", tr[v[i]].front().second - 1);
      tr[v[i]].pop_front();
    }
  return 0;
}

```