#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5, OO = 1e12;
vector<long long> son[N], v, h[N];
long long dp[N][64], siz[N], str[N], fin[N], t = 1;
void dfs(long long x, long long z, long long y) {
  str[x] = t++;
  siz[x] = z;
  h[z].push_back(str[x]);
  dp[x][0] = y;
  for (int i = 1; i <= 30; i++) dp[x][i] = dp[dp[x][i - 1]][i - 1];
  for (auto u : son[x])
    if (u != y) dfs(u, z + 1, x);
  fin[x] = t++;
  h[z].push_back(fin[x]);
}
int main() {
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    if (x == 0)
      v.push_back(i);
    else
      son[x].push_back(i);
  }
  for (auto u : v) dfs(u, 0, 0);
  long long q;
  cin >> q;
  vector<long long> vw;
  while (q--) {
    long long x, k, f;
    cin >> x >> k;
    f = k;
    vector<long long> v1;
    while (f > 0) {
      v1.push_back(f % 2);
      f /= 2;
    }
    long long g = x;
    for (int i = 0; i < v1.size(); i++) {
      if (v1[i] == 1) g = dp[g][i];
    }
    if (g == 0)
      vw.push_back(0);
    else {
      f = siz[g] + k;
      long long s1 =
          lower_bound(h[f].begin(), h[f].end(), str[g]) - h[f].begin();
      long long f1 =
          lower_bound(h[f].begin(), h[f].end(), fin[g]) - h[f].begin();
      vw.push_back((f1 - s1) / 2 - 1);
    }
  }
  for (auto u : vw) cout << u << " ";
  return 0;
}

### Prompt

Please create a solution in Cpp 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 long long N = 2e5, OO = 1e12;
vector<long long> son[N], v, h[N];
long long dp[N][64], siz[N], str[N], fin[N], t = 1;
void dfs(long long x, long long z, long long y) {
  str[x] = t++;
  siz[x] = z;
  h[z].push_back(str[x]);
  dp[x][0] = y;
  for (int i = 1; i <= 30; i++) dp[x][i] = dp[dp[x][i - 1]][i - 1];
  for (auto u : son[x])
    if (u != y) dfs(u, z + 1, x);
  fin[x] = t++;
  h[z].push_back(fin[x]);
}
int main() {
  long long n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    if (x == 0)
      v.push_back(i);
    else
      son[x].push_back(i);
  }
  for (auto u : v) dfs(u, 0, 0);
  long long q;
  cin >> q;
  vector<long long> vw;
  while (q--) {
    long long x, k, f;
    cin >> x >> k;
    f = k;
    vector<long long> v1;
    while (f > 0) {
      v1.push_back(f % 2);
      f /= 2;
    }
    long long g = x;
    for (int i = 0; i < v1.size(); i++) {
      if (v1[i] == 1) g = dp[g][i];
    }
    if (g == 0)
      vw.push_back(0);
    else {
      f = siz[g] + k;
      long long s1 =
          lower_bound(h[f].begin(), h[f].end(), str[g]) - h[f].begin();
      long long f1 =
          lower_bound(h[f].begin(), h[f].end(), fin[g]) - h[f].begin();
      vw.push_back((f1 - s1) / 2 - 1);
    }
  }
  for (auto u : vw) cout << u << " ";
  return 0;
}

```