#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000 * 100 + 100;
vector<int> adj[MAXN], d[MAXN], path;
int start[MAXN], fin[MAXN], deep[MAXN], cnt = 1, par[MAXN][21];
void dfs(int dd, int u) {
  for (int i = 0; i < 21; i++) {
    int x = (1 << i);
    if (x <= path.size()) par[u][i] = path[path.size() - x];
  }
  path.push_back(u);
  deep[u] = dd;
  start[u] = cnt++;
  d[dd].push_back(start[u]);
  for (int i = 0; i < adj[u].size(); i++) dfs(dd + 1, adj[u][i]);
  path.pop_back();
  fin[u] = cnt++;
}
int find_par(int a, int b) {
  if (b == 0) return a;
  int x = 0;
  for (int i = 0; i < 20; i++)
    if ((1 << i) <= b and (1 << (i + 1)) > b)
      return find_par(par[a][i], b - (1 << i));
}
int find(int r, int dd) {
  if (r == 0) return 1;
  int up = upper_bound(d[dd].begin(), d[dd].end(), fin[r]) - d[dd].begin();
  int down = lower_bound(d[dd].begin(), d[dd].end(), start[r]) - d[dd].begin();
  return up - down;
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int r;
    scanf("%d", &r);
    adj[r].push_back(i);
  }
  dfs(0, 0);
  for (int i = 0; i <= n; i++) sort(d[i].begin(), d[i].end());
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    int p = find_par(a, b);
    int x = find(p, deep[a]) - 1;
    printf("%d ", x);
  }
  printf("\n");
  return 0;
}

### Prompt

In cpp, your task is to solve 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 MAXN = 1000 * 100 + 100;
vector<int> adj[MAXN], d[MAXN], path;
int start[MAXN], fin[MAXN], deep[MAXN], cnt = 1, par[MAXN][21];
void dfs(int dd, int u) {
  for (int i = 0; i < 21; i++) {
    int x = (1 << i);
    if (x <= path.size()) par[u][i] = path[path.size() - x];
  }
  path.push_back(u);
  deep[u] = dd;
  start[u] = cnt++;
  d[dd].push_back(start[u]);
  for (int i = 0; i < adj[u].size(); i++) dfs(dd + 1, adj[u][i]);
  path.pop_back();
  fin[u] = cnt++;
}
int find_par(int a, int b) {
  if (b == 0) return a;
  int x = 0;
  for (int i = 0; i < 20; i++)
    if ((1 << i) <= b and (1 << (i + 1)) > b)
      return find_par(par[a][i], b - (1 << i));
}
int find(int r, int dd) {
  if (r == 0) return 1;
  int up = upper_bound(d[dd].begin(), d[dd].end(), fin[r]) - d[dd].begin();
  int down = lower_bound(d[dd].begin(), d[dd].end(), start[r]) - d[dd].begin();
  return up - down;
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int r;
    scanf("%d", &r);
    adj[r].push_back(i);
  }
  dfs(0, 0);
  for (int i = 0; i <= n; i++) sort(d[i].begin(), d[i].end());
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    int p = find_par(a, b);
    int x = find(p, deep[a]) - 1;
    printf("%d ", x);
  }
  printf("\n");
  return 0;
}

```