#include <bits/stdc++.h>
using namespace std;
struct Tree {
  vector<vector<int> > adj, up, lay;
  int cnt;
  vector<int> pre, post, lvl, cur;
  Tree(int n)
      : adj(n + 1),
        up(n + 1),
        cnt(0),
        lay(1),
        pre(n + 1),
        post(n + 1),
        lvl(n + 1),
        cur(n + 1) {}
  void addEdge(int u, int v) {
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  void dfs(int u, int par) {
    pre[u] = cnt++;
    for (int v : adj[u]) {
      if (v != par) {
        lvl[v] = 1 + lvl[u];
        if (lvl[v] == lay.size()) {
          lay.push_back(vector<int>(1, v));
        } else {
          lay[lvl[v]].push_back(v);
        }
        cur[v] = lay[lvl[v]].size() - 1;
        int a = u;
        up[v].push_back(a);
        for (int i = 0; i < up[a].size(); i++) {
          a = up[a][i];
          up[v].push_back(a);
        }
        dfs(v, u);
      }
    }
    post[u] = cnt++;
  }
  void build(int n) {
    for (int r = 1; r <= n; r++) {
      if (pre[r] == 0) {
        lay[0].push_back(r);
        dfs(r, r);
      }
    }
  }
  bool ancestor(int u, int v) {
    return (pre[u] <= pre[v] && post[u] >= post[v]);
  }
  int lift(int u, int k) {
    for (int i = 0; k; i++) {
      if (i >= up[u].size()) {
        throw -1;
      } else {
        if (k & 1) u = up[u][i];
        k >>= 1;
      }
    }
    return u;
  }
  int cousins(int u, int k) {
    try {
      int a = lift(u, k);
      int l = 0, r = cur[u];
      while (l < r) {
        int m = (l + r) >> 1;
        if (ancestor(a, lay[lvl[u]][m])) {
          r = m;
        } else {
          l = m + 1;
        }
      }
      int total = cur[u] - l;
      l = cur[u], r = lay[lvl[u]].size();
      while (r - l > 1) {
        int m = (l + r) >> 1;
        if (ancestor(a, lay[lvl[u]][m])) {
          l = m;
        } else {
          r = m;
        }
      }
      total += l - cur[u];
      return total;
    } catch (int e) {
      return 0;
    }
  }
};
int main() {
  int n;
  scanf("%d", &n);
  Tree t(n);
  for (int u = 1; u <= n; u++) {
    int par;
    scanf("%d", &par);
    if (par > 0) {
      t.addEdge(u, par);
    }
  }
  t.build(n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int u, k;
    scanf("%d %d", &u, &k);
    printf("%d ", t.cousins(u, k));
  }
}

### 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;
struct Tree {
  vector<vector<int> > adj, up, lay;
  int cnt;
  vector<int> pre, post, lvl, cur;
  Tree(int n)
      : adj(n + 1),
        up(n + 1),
        cnt(0),
        lay(1),
        pre(n + 1),
        post(n + 1),
        lvl(n + 1),
        cur(n + 1) {}
  void addEdge(int u, int v) {
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  void dfs(int u, int par) {
    pre[u] = cnt++;
    for (int v : adj[u]) {
      if (v != par) {
        lvl[v] = 1 + lvl[u];
        if (lvl[v] == lay.size()) {
          lay.push_back(vector<int>(1, v));
        } else {
          lay[lvl[v]].push_back(v);
        }
        cur[v] = lay[lvl[v]].size() - 1;
        int a = u;
        up[v].push_back(a);
        for (int i = 0; i < up[a].size(); i++) {
          a = up[a][i];
          up[v].push_back(a);
        }
        dfs(v, u);
      }
    }
    post[u] = cnt++;
  }
  void build(int n) {
    for (int r = 1; r <= n; r++) {
      if (pre[r] == 0) {
        lay[0].push_back(r);
        dfs(r, r);
      }
    }
  }
  bool ancestor(int u, int v) {
    return (pre[u] <= pre[v] && post[u] >= post[v]);
  }
  int lift(int u, int k) {
    for (int i = 0; k; i++) {
      if (i >= up[u].size()) {
        throw -1;
      } else {
        if (k & 1) u = up[u][i];
        k >>= 1;
      }
    }
    return u;
  }
  int cousins(int u, int k) {
    try {
      int a = lift(u, k);
      int l = 0, r = cur[u];
      while (l < r) {
        int m = (l + r) >> 1;
        if (ancestor(a, lay[lvl[u]][m])) {
          r = m;
        } else {
          l = m + 1;
        }
      }
      int total = cur[u] - l;
      l = cur[u], r = lay[lvl[u]].size();
      while (r - l > 1) {
        int m = (l + r) >> 1;
        if (ancestor(a, lay[lvl[u]][m])) {
          l = m;
        } else {
          r = m;
        }
      }
      total += l - cur[u];
      return total;
    } catch (int e) {
      return 0;
    }
  }
};
int main() {
  int n;
  scanf("%d", &n);
  Tree t(n);
  for (int u = 1; u <= n; u++) {
    int par;
    scanf("%d", &par);
    if (par > 0) {
      t.addEdge(u, par);
    }
  }
  t.build(n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int u, k;
    scanf("%d %d", &u, &k);
    printf("%d ", t.cousins(u, k));
  }
}

```