#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > adj, level;
int par[100001];
int in[100001], out[100001], rnk[100001];
int timer, n;
void dfs(int i, int depth) {
  in[i] = timer++;
  rnk[i] = depth;
  level[depth].push_back(in[i]);
  for (int j = 0; j < adj[i].size(); j++) {
    int next = adj[i][j];
    dfs(next, depth + 1);
  }
  out[i] = timer++;
}
int sparse[100001][32];
void build() {
  for (int i = 0; i < n; i++) {
    sparse[i][0] = par[i];
  }
  for (int i = 1; i < 32; i++) {
    for (int j = 0; j < n; j++) {
      if (sparse[j][i - 1] != -1) {
        sparse[j][i] = sparse[sparse[j][i - 1]][i - 1];
      }
    }
  }
}
int pth(int node, int p) {
  int tmp = 30;
  while (p > 0 && node >= 0) {
    while ((1 << tmp) > p) {
      tmp--;
    }
    node = sparse[node][tmp];
    p -= (1 << tmp);
  }
  return node;
}
int main() {
  while (cin >> n) {
    timer = 0;
    memset(par, -1, sizeof(par));
    memset(sparse, -1, sizeof(sparse));
    adj.clear();
    adj.resize(n);
    level.clear();
    level.resize(n);
    int u;
    for (int i = 0; i < n; i++) {
      cin >> u;
      if (u) {
        par[i] = u - 1;
        adj[u - 1].push_back(i);
      }
    }
    for (int i = 0; i < n; i++) {
      if (par[i] == -1) {
        dfs(i, 0);
      }
    }
    for (int i = 0; i < n; i++) {
      sort(level[i].begin(), level[i].end());
    }
    build();
    int m;
    cin >> m;
    int p;
    while (m--) {
      cin >> u >> p;
      u--;
      int parent = pth(u, p);
      if (parent == -1) {
        cout << 0 << endl;
      } else {
        cout << lower_bound(level[rnk[u]].begin(), level[rnk[u]].end(),
                            out[parent]) -
                    lower_bound(level[rnk[u]].begin(), level[rnk[u]].end(),
                                in[parent]) -
                    1
             << endl;
      }
    }
  }
  return 0;
}

### 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<vector<int> > adj, level;
int par[100001];
int in[100001], out[100001], rnk[100001];
int timer, n;
void dfs(int i, int depth) {
  in[i] = timer++;
  rnk[i] = depth;
  level[depth].push_back(in[i]);
  for (int j = 0; j < adj[i].size(); j++) {
    int next = adj[i][j];
    dfs(next, depth + 1);
  }
  out[i] = timer++;
}
int sparse[100001][32];
void build() {
  for (int i = 0; i < n; i++) {
    sparse[i][0] = par[i];
  }
  for (int i = 1; i < 32; i++) {
    for (int j = 0; j < n; j++) {
      if (sparse[j][i - 1] != -1) {
        sparse[j][i] = sparse[sparse[j][i - 1]][i - 1];
      }
    }
  }
}
int pth(int node, int p) {
  int tmp = 30;
  while (p > 0 && node >= 0) {
    while ((1 << tmp) > p) {
      tmp--;
    }
    node = sparse[node][tmp];
    p -= (1 << tmp);
  }
  return node;
}
int main() {
  while (cin >> n) {
    timer = 0;
    memset(par, -1, sizeof(par));
    memset(sparse, -1, sizeof(sparse));
    adj.clear();
    adj.resize(n);
    level.clear();
    level.resize(n);
    int u;
    for (int i = 0; i < n; i++) {
      cin >> u;
      if (u) {
        par[i] = u - 1;
        adj[u - 1].push_back(i);
      }
    }
    for (int i = 0; i < n; i++) {
      if (par[i] == -1) {
        dfs(i, 0);
      }
    }
    for (int i = 0; i < n; i++) {
      sort(level[i].begin(), level[i].end());
    }
    build();
    int m;
    cin >> m;
    int p;
    while (m--) {
      cin >> u >> p;
      u--;
      int parent = pth(u, p);
      if (parent == -1) {
        cout << 0 << endl;
      } else {
        cout << lower_bound(level[rnk[u]].begin(), level[rnk[u]].end(),
                            out[parent]) -
                    lower_bound(level[rnk[u]].begin(), level[rnk[u]].end(),
                                in[parent]) -
                    1
             << endl;
      }
    }
  }
  return 0;
}

```