#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const int l = 20;
int up[maxn][l];
vector<int> a[maxn];
int in[maxn], out[maxn], lvl[maxn];
int cnt, n;
vector<int> dist[maxn];
int ans[maxn];
void dfs(int u, int p) {
  if (u) {
    in[u] = ++cnt;
    lvl[u] = lvl[p] + 1;
    dist[lvl[u]].push_back(in[u]);
  }
  up[u][0] = p;
  for (int i = 1; i < l; ++i) up[u][i] = up[up[u][i - 1]][i - 1];
  for (auto &v : a[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
  out[u] = cnt;
}
int get_kth_ancestor(int x, int k) {
  for (int i = 0; i < l; i++) {
    if (k & (1 << i)) x = up[x][i];
  }
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[x].push_back(i);
    a[i].push_back(x);
  }
  dfs(0, 0);
  int m;
  cin >> m;
  for (int q = 1; q <= m; q++) {
    int v, p;
    cin >> v >> p;
    int x = get_kth_ancestor(v, p);
    if (x == 0) {
      ans[q] = 0;
      continue;
    }
    int up = lower_bound(dist[lvl[v]].begin(), dist[lvl[v]].end(), out[x] + 1) -
             dist[lvl[v]].begin();
    int lo = lower_bound(dist[lvl[v]].begin(), dist[lvl[v]].end(), in[x]) -
             dist[lvl[v]].begin();
    ans[q] = up - lo - 1;
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << " ";
  return 0;
}

### Prompt

Your challenge is to write 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;
const int maxn = 100005;
const int l = 20;
int up[maxn][l];
vector<int> a[maxn];
int in[maxn], out[maxn], lvl[maxn];
int cnt, n;
vector<int> dist[maxn];
int ans[maxn];
void dfs(int u, int p) {
  if (u) {
    in[u] = ++cnt;
    lvl[u] = lvl[p] + 1;
    dist[lvl[u]].push_back(in[u]);
  }
  up[u][0] = p;
  for (int i = 1; i < l; ++i) up[u][i] = up[up[u][i - 1]][i - 1];
  for (auto &v : a[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
  out[u] = cnt;
}
int get_kth_ancestor(int x, int k) {
  for (int i = 0; i < l; i++) {
    if (k & (1 << i)) x = up[x][i];
  }
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    a[x].push_back(i);
    a[i].push_back(x);
  }
  dfs(0, 0);
  int m;
  cin >> m;
  for (int q = 1; q <= m; q++) {
    int v, p;
    cin >> v >> p;
    int x = get_kth_ancestor(v, p);
    if (x == 0) {
      ans[q] = 0;
      continue;
    }
    int up = lower_bound(dist[lvl[v]].begin(), dist[lvl[v]].end(), out[x] + 1) -
             dist[lvl[v]].begin();
    int lo = lower_bound(dist[lvl[v]].begin(), dist[lvl[v]].end(), in[x]) -
             dist[lvl[v]].begin();
    ans[q] = up - lo - 1;
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << " ";
  return 0;
}

```