#include <bits/stdc++.h>
using namespace std;
const int MAX_N = int(1e5) + 7;
int f[MAX_N][18];
vector<int> G[MAX_N];
int b[MAX_N], e[MAX_N], lvl[MAX_N], t;
vector<int> lv[MAX_N];
void dfs(int x, int l) {
  b[x] = ++t;
  lvl[x] = l;
  lv[l].push_back(b[x]);
  for (int i = 0; i < G[x].size(); i++) dfs(G[x][i], l + 1);
  e[x] = t;
}
int main() {
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> f[i][0];
    G[f[i][0]].push_back(i);
  }
  t = -1;
  dfs(0, 0);
  for (int j = 1; j <= 17; j++)
    for (int i = 1; i <= n; i++) f[i][j] = f[f[i][j - 1]][j - 1];
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int p, v, l;
    cin >> v >> p;
    l = lvl[v];
    for (int i = 0; i <= 17; i++)
      if (p & (1 << i)) v = f[v][i];
    if (v) {
      int ans = lower_bound(lv[l].begin(), lv[l].end(), e[v] + 1) -
                lower_bound(lv[l].begin(), lv[l].end(), b[v]) - 1;
      cout << ans << " ";
    } else
      cout << "0 ";
  }
  cout << 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;
const int MAX_N = int(1e5) + 7;
int f[MAX_N][18];
vector<int> G[MAX_N];
int b[MAX_N], e[MAX_N], lvl[MAX_N], t;
vector<int> lv[MAX_N];
void dfs(int x, int l) {
  b[x] = ++t;
  lvl[x] = l;
  lv[l].push_back(b[x]);
  for (int i = 0; i < G[x].size(); i++) dfs(G[x][i], l + 1);
  e[x] = t;
}
int main() {
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> f[i][0];
    G[f[i][0]].push_back(i);
  }
  t = -1;
  dfs(0, 0);
  for (int j = 1; j <= 17; j++)
    for (int i = 1; i <= n; i++) f[i][j] = f[f[i][j - 1]][j - 1];
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int p, v, l;
    cin >> v >> p;
    l = lvl[v];
    for (int i = 0; i <= 17; i++)
      if (p & (1 << i)) v = f[v][i];
    if (v) {
      int ans = lower_bound(lv[l].begin(), lv[l].end(), e[v] + 1) -
                lower_bound(lv[l].begin(), lv[l].end(), b[v]) - 1;
      cout << ans << " ";
    } else
      cout << "0 ";
  }
  cout << endl;
  return 0;
}

```