#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 3;
const int maxm = 1e7 + 3;
const int mod = 1e9 + 7;
const int oo = 1e9 + 5;
const long double pi = 3.141592653589793238462643383279502884;
void input() {
  if (fopen("abc"
            ".inp",
            "r")) {
    freopen(
        "abc"
        ".inp",
        "r", stdin);
    freopen(
        "abc"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int n, m, _time = 0;
vector<int> g[maxn], val[maxn];
int in[maxn], out[maxn], anc[maxn][20], h[maxn];
void dfs(int u) {
  in[u] = ++_time;
  val[h[u]].push_back(in[u]);
  for (int v : g[u]) {
    h[v] = h[u] + 1;
    dfs(v);
  }
  out[u] = _time;
}
int jump(int u, int d) {
  for (int i = 18; i >= 0; --i)
    if ((d >> i) & 1) u = anc[u][i];
  return u;
}
int main() {
  input();
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> anc[i][0];
    g[anc[i][0]].push_back(i);
  }
  for (int j = 1; j < 19; ++j)
    for (int i = 1; i <= n; ++i) anc[i][j] = anc[anc[i][j - 1]][j - 1];
  dfs(0);
  cin >> m;
  int v, p;
  while (m--) {
    cin >> v >> p;
    int u = jump(v, p);
    if (!u)
      cout << "0\n";
    else {
      auto l = lower_bound(val[h[v]].begin(), val[h[v]].end(), in[u]) -
               val[h[v]].begin();
      auto r = upper_bound(val[h[v]].begin(), val[h[v]].end(), out[u]) -
               val[h[v]].begin() - 1;
      cout << r - l << "\n";
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 maxn = 2e5 + 3;
const int maxm = 1e7 + 3;
const int mod = 1e9 + 7;
const int oo = 1e9 + 5;
const long double pi = 3.141592653589793238462643383279502884;
void input() {
  if (fopen("abc"
            ".inp",
            "r")) {
    freopen(
        "abc"
        ".inp",
        "r", stdin);
    freopen(
        "abc"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int n, m, _time = 0;
vector<int> g[maxn], val[maxn];
int in[maxn], out[maxn], anc[maxn][20], h[maxn];
void dfs(int u) {
  in[u] = ++_time;
  val[h[u]].push_back(in[u]);
  for (int v : g[u]) {
    h[v] = h[u] + 1;
    dfs(v);
  }
  out[u] = _time;
}
int jump(int u, int d) {
  for (int i = 18; i >= 0; --i)
    if ((d >> i) & 1) u = anc[u][i];
  return u;
}
int main() {
  input();
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> anc[i][0];
    g[anc[i][0]].push_back(i);
  }
  for (int j = 1; j < 19; ++j)
    for (int i = 1; i <= n; ++i) anc[i][j] = anc[anc[i][j - 1]][j - 1];
  dfs(0);
  cin >> m;
  int v, p;
  while (m--) {
    cin >> v >> p;
    int u = jump(v, p);
    if (!u)
      cout << "0\n";
    else {
      auto l = lower_bound(val[h[v]].begin(), val[h[v]].end(), in[u]) -
               val[h[v]].begin();
      auto r = upper_bound(val[h[v]].begin(), val[h[v]].end(), out[u]) -
               val[h[v]].begin() - 1;
      cout << r - l << "\n";
    }
  }
  return 0;
}

```