#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
vector<int> V[N], G[N];
int cnt = 0;
int In[N], Out[N], dp[N];
int par[N][20];
void dfs(int idx, int sum) {
  Out[idx] = In[idx] = ++cnt;
  dp[idx] = sum;
  G[sum].push_back(cnt);
  for (auto k : V[idx]) {
    dfs(k, sum + 1);
    Out[idx] = ++cnt;
  }
}
int main() {
  int n, x, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", &x), par[i][0] = x, V[x].push_back(i);
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= n; j++) {
      if (par[j][i - 1] != 0 && par[par[j][i - 1]][i - 1] != 0)
        par[j][i] = par[par[j][i - 1]][i - 1];
    }
  }
  for (int i = 1; i <= n; i++)
    if (!par[i][0]) dfs(i, 1);
  scanf("%d", &q);
  while (q--) {
    int v, p;
    scanf("%d%d", &v, &p);
    int g = p, j = v, cc = 0;
    while (g) {
      if (g & 1) j = par[j][cc];
      g >>= 1;
      cc++;
    }
    if (j == 0)
      cout << 0 << '\n';
    else {
      int l = In[j], r = Out[j];
      vector<int>& f = G[dp[v]];
      cout << lower_bound(f.begin(), f.end(), r) -
                  lower_bound(f.begin(), f.end(), l) - 1
           << '\n';
    }
  }
}

### Prompt

Generate 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 N = 1e5 + 10;
vector<int> V[N], G[N];
int cnt = 0;
int In[N], Out[N], dp[N];
int par[N][20];
void dfs(int idx, int sum) {
  Out[idx] = In[idx] = ++cnt;
  dp[idx] = sum;
  G[sum].push_back(cnt);
  for (auto k : V[idx]) {
    dfs(k, sum + 1);
    Out[idx] = ++cnt;
  }
}
int main() {
  int n, x, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", &x), par[i][0] = x, V[x].push_back(i);
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= n; j++) {
      if (par[j][i - 1] != 0 && par[par[j][i - 1]][i - 1] != 0)
        par[j][i] = par[par[j][i - 1]][i - 1];
    }
  }
  for (int i = 1; i <= n; i++)
    if (!par[i][0]) dfs(i, 1);
  scanf("%d", &q);
  while (q--) {
    int v, p;
    scanf("%d%d", &v, &p);
    int g = p, j = v, cc = 0;
    while (g) {
      if (g & 1) j = par[j][cc];
      g >>= 1;
      cc++;
    }
    if (j == 0)
      cout << 0 << '\n';
    else {
      int l = In[j], r = Out[j];
      vector<int>& f = G[dp[v]];
      cout << lower_bound(f.begin(), f.end(), r) -
                  lower_bound(f.begin(), f.end(), l) - 1
           << '\n';
    }
  }
}

```