#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m;
int anc[N][20];
vector<int> g[N];
int t, ans;
int in_time[N];
int out_time[N];
int deep[N];
vector<int> v[N];
void dfs(int i, int p) {
  in_time[i] = t;
  v[deep[i]].push_back(t);
  t++;
  for (auto x : g[i]) {
    if (x == p) continue;
    deep[x] = deep[i] + 1;
    dfs(x, i);
  }
  out_time[i] = t++;
}
void lca_in_it() {
  for (int j = 1; j < 20; j++) {
    for (int i = 1; i <= n; i++) {
      int last = anc[i][j - 1];
      anc[i][j] = anc[last][j - 1];
    }
  }
}
int query(int x, int p) {
  if (deep[x] <= p) return 0;
  int level = deep[x];
  for (int jump = 19; jump >= 0; jump--) {
    if (p >= (1 << jump)) {
      p = p - (1 << jump);
      x = anc[x][jump];
    }
  }
  int lo = upper_bound(v[level].begin(), v[level].end(), in_time[x]) -
           v[level].begin();
  int up = lower_bound(v[level].begin(), v[level].end(), out_time[x] - 1) -
           v[level].begin();
  return (up - lo - 1);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    anc[i][0] = x;
    g[x].push_back(i);
  }
  dfs(0, -1);
  lca_in_it();
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    ans = query(x, y);
    cout << ans << " ";
  }
}

### 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;
const int N = 1e5 + 10;
int n, m;
int anc[N][20];
vector<int> g[N];
int t, ans;
int in_time[N];
int out_time[N];
int deep[N];
vector<int> v[N];
void dfs(int i, int p) {
  in_time[i] = t;
  v[deep[i]].push_back(t);
  t++;
  for (auto x : g[i]) {
    if (x == p) continue;
    deep[x] = deep[i] + 1;
    dfs(x, i);
  }
  out_time[i] = t++;
}
void lca_in_it() {
  for (int j = 1; j < 20; j++) {
    for (int i = 1; i <= n; i++) {
      int last = anc[i][j - 1];
      anc[i][j] = anc[last][j - 1];
    }
  }
}
int query(int x, int p) {
  if (deep[x] <= p) return 0;
  int level = deep[x];
  for (int jump = 19; jump >= 0; jump--) {
    if (p >= (1 << jump)) {
      p = p - (1 << jump);
      x = anc[x][jump];
    }
  }
  int lo = upper_bound(v[level].begin(), v[level].end(), in_time[x]) -
           v[level].begin();
  int up = lower_bound(v[level].begin(), v[level].end(), out_time[x] - 1) -
           v[level].begin();
  return (up - lo - 1);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    anc[i][0] = x;
    g[x].push_back(i);
  }
  dfs(0, -1);
  lca_in_it();
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    ans = query(x, y);
    cout << ans << " ";
  }
}

```