#include <bits/stdc++.h>
using namespace std;
vector<int> g[100002];
int r[100002];
int s[100002], e[100002];
int T = 1;
int par[17][100002], depth[100002];
vector<int> d[100002];
void dfs(int u, int p, int d) {
  s[u] = T++;
  par[0][u] = p;
  depth[u] = d;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    dfs(v, u, d + 1);
  }
  e[u] = T - 1;
}
int jump(int u, int k) {
  if (depth[u] < k) return -1;
  for (int i = 16; i >= 0; i--) {
    if ((k >> i) & 1) u = par[i][u];
  }
  return u;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &r[i]);
    if (r[i] != 0) {
      g[r[i]].emplace_back(i);
      g[i].emplace_back(r[i]);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (r[i] == 0) dfs(i, 0, 0);
  }
  for (int i = 1; i <= n; i++) {
    d[depth[i]].emplace_back(s[i]);
  }
  for (int i = 0; i <= n; i++) sort(d[i].begin(), d[i].end());
  for (int j = 1; j < 17; j++) {
    for (int i = 1; i <= n; i++) {
      if (par[j - 1][i] != 0) par[j][i] = par[j - 1][par[j - 1][i]];
    }
  }
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int v, p;
    scanf("%d%d", &v, &p);
    int u = jump(v, p);
    if (u == -1)
      printf("0 ");
    else {
      int D = depth[u] + p;
      int l = lower_bound(d[D].begin(), d[D].end(), s[u]) - d[D].begin();
      int r = upper_bound(d[D].begin(), d[D].end(), e[u]) - d[D].begin() - 1;
      printf("%d ", r - l + 1 - 1);
    }
  }
}

### 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;
vector<int> g[100002];
int r[100002];
int s[100002], e[100002];
int T = 1;
int par[17][100002], depth[100002];
vector<int> d[100002];
void dfs(int u, int p, int d) {
  s[u] = T++;
  par[0][u] = p;
  depth[u] = d;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    dfs(v, u, d + 1);
  }
  e[u] = T - 1;
}
int jump(int u, int k) {
  if (depth[u] < k) return -1;
  for (int i = 16; i >= 0; i--) {
    if ((k >> i) & 1) u = par[i][u];
  }
  return u;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &r[i]);
    if (r[i] != 0) {
      g[r[i]].emplace_back(i);
      g[i].emplace_back(r[i]);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (r[i] == 0) dfs(i, 0, 0);
  }
  for (int i = 1; i <= n; i++) {
    d[depth[i]].emplace_back(s[i]);
  }
  for (int i = 0; i <= n; i++) sort(d[i].begin(), d[i].end());
  for (int j = 1; j < 17; j++) {
    for (int i = 1; i <= n; i++) {
      if (par[j - 1][i] != 0) par[j][i] = par[j - 1][par[j - 1][i]];
    }
  }
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int v, p;
    scanf("%d%d", &v, &p);
    int u = jump(v, p);
    if (u == -1)
      printf("0 ");
    else {
      int D = depth[u] + p;
      int l = lower_bound(d[D].begin(), d[D].end(), s[u]) - d[D].begin();
      int r = upper_bound(d[D].begin(), d[D].end(), e[u]) - d[D].begin() - 1;
      printf("%d ", r - l + 1 - 1);
    }
  }
}

```