#include <bits/stdc++.h>
using namespace std;
const int SIZE = 100002, LOG = 18;
vector<int> root;
vector<int> g[SIZE];
int g1[SIZE];
vector<pair<int, int> > q[SIZE];
int h[SIZE + 5];
int d[SIZE + 5];
int tin[SIZE], tout[SIZE], t[SIZE], tim = -1, s[SIZE];
int ans[SIZE];
int p[SIZE][LOG];
void calc(int v, int e) {
  h[v] = e;
  tim++;
  tin[v] = tim;
  t[tim] = v;
  s[v] = 1;
  p[v][0] = g1[v];
  for (int i = 1; i < LOG; i++) {
    if (p[v][i - 1] != -1) {
      p[v][i] = p[p[v][i - 1]][i - 1];
    } else {
      p[v][i] = -1;
    }
  }
  for (int u : g[v]) {
    calc(u, e + 1);
    s[v] += s[u];
  }
  tout[v] = tim;
}
void dfs(int v, bool f) {
  int maxv = -1, maxr = 0;
  for (int u : g[v]) {
    if (s[u] > maxr) {
      maxr = s[u];
      maxv = u;
    }
  }
  for (int u : g[v]) {
    if (u != maxv) {
      dfs(u, 0);
    }
  }
  if (maxv != -1) {
    dfs(maxv, 1);
  }
  for (int u : g[v]) {
    if (u == maxv) {
      continue;
    }
    for (int w = tin[u]; w <= tout[u]; w++) {
      int y = t[w];
      d[h[y]]++;
    }
  }
  d[h[v]]++;
  for (auto w1 : q[v]) {
    int w = w1.first;
    int ans1 = h[v] + w;
    if (ans1 >= SIZE) {
      ans[w1.second] = 0;
    } else {
      ans[w1.second] = d[h[v] + w] - 1;
    }
  }
  if (f == 0) {
    for (int w = tin[v]; w <= tout[v]; w++) {
      int y = t[w];
      d[h[y]]--;
    }
  }
}
int up(int v, int e) {
  for (int i = LOG - 1; i > -1; i--) {
    if (v == -1) {
      return -1;
    }
    if (((1 << i) & e) != 0) {
      v = p[v][i];
    }
  }
  return v;
}
int main() {
  int n, m, a, b;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a == 0) {
      root.push_back(i);
      g1[i] = -1;
    } else {
      g[a - 1].push_back(i);
      g1[i] = a - 1;
    }
  }
  cin >> m;
  for (int w : root) {
    calc(w, 1);
  }
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    int v = up(a - 1, b);
    if (v == -1) {
      continue;
    }
    q[v].push_back({b, i});
  }
  for (int w : root) {
    dfs(w, 0);
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 SIZE = 100002, LOG = 18;
vector<int> root;
vector<int> g[SIZE];
int g1[SIZE];
vector<pair<int, int> > q[SIZE];
int h[SIZE + 5];
int d[SIZE + 5];
int tin[SIZE], tout[SIZE], t[SIZE], tim = -1, s[SIZE];
int ans[SIZE];
int p[SIZE][LOG];
void calc(int v, int e) {
  h[v] = e;
  tim++;
  tin[v] = tim;
  t[tim] = v;
  s[v] = 1;
  p[v][0] = g1[v];
  for (int i = 1; i < LOG; i++) {
    if (p[v][i - 1] != -1) {
      p[v][i] = p[p[v][i - 1]][i - 1];
    } else {
      p[v][i] = -1;
    }
  }
  for (int u : g[v]) {
    calc(u, e + 1);
    s[v] += s[u];
  }
  tout[v] = tim;
}
void dfs(int v, bool f) {
  int maxv = -1, maxr = 0;
  for (int u : g[v]) {
    if (s[u] > maxr) {
      maxr = s[u];
      maxv = u;
    }
  }
  for (int u : g[v]) {
    if (u != maxv) {
      dfs(u, 0);
    }
  }
  if (maxv != -1) {
    dfs(maxv, 1);
  }
  for (int u : g[v]) {
    if (u == maxv) {
      continue;
    }
    for (int w = tin[u]; w <= tout[u]; w++) {
      int y = t[w];
      d[h[y]]++;
    }
  }
  d[h[v]]++;
  for (auto w1 : q[v]) {
    int w = w1.first;
    int ans1 = h[v] + w;
    if (ans1 >= SIZE) {
      ans[w1.second] = 0;
    } else {
      ans[w1.second] = d[h[v] + w] - 1;
    }
  }
  if (f == 0) {
    for (int w = tin[v]; w <= tout[v]; w++) {
      int y = t[w];
      d[h[y]]--;
    }
  }
}
int up(int v, int e) {
  for (int i = LOG - 1; i > -1; i--) {
    if (v == -1) {
      return -1;
    }
    if (((1 << i) & e) != 0) {
      v = p[v][i];
    }
  }
  return v;
}
int main() {
  int n, m, a, b;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a == 0) {
      root.push_back(i);
      g1[i] = -1;
    } else {
      g[a - 1].push_back(i);
      g1[i] = a - 1;
    }
  }
  cin >> m;
  for (int w : root) {
    calc(w, 1);
  }
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    int v = up(a - 1, b);
    if (v == -1) {
      continue;
    }
    q[v].push_back({b, i});
  }
  for (int w : root) {
    dfs(w, 0);
  }
  for (int i = 0; i < m; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}

```