#include <bits/stdc++.h>
using namespace std;
int n, m, h[100015], p[20][100015], dfsCnt, num[100015], cnt[100015];
vector<int> a[100015], s[100015];
void dfsLCA(int u) {
  num[u] = ++dfsCnt;
  cnt[u] = 1;
  s[h[u]].push_back(num[u]);
  for (int v : a[u]) h[v] = h[u] + 1, dfsLCA(v), cnt[u] += cnt[v];
}
void initLCA() {
  for (int u = 1; u <= n; ++u)
    if (!p[0][u]) h[u] = 1, dfsLCA(u);
  for (int k = 1, p2k = 2; p2k <= n; ++k, p2k <<= 1)
    for (int u = 1; u <= n; ++u) p[k][u] = p[k - 1][p[k - 1][u]];
}
int ancestor(int u, int x) {
  for (int k; x; x ^= 1 << k) k = log2(x), u = p[k][u];
  return u;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> p[0][i], a[p[0][i]].push_back(i);
  initLCA();
  for (int i = 1; i <= n; ++i) sort(s[i].begin(), s[i].end());
  cin >> m;
  for (int v, k, x; m--;) {
    cin >> v >> k;
    x = ancestor(v, k);
    if (!x)
      cout << "0 ";
    else
      cout << lower_bound(s[h[v]].begin(), s[h[v]].end(), num[x] + cnt[x]) -
                  lower_bound(s[h[v]].begin(), s[h[v]].end(), num[x]) - 1
           << ' ';
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int n, m, h[100015], p[20][100015], dfsCnt, num[100015], cnt[100015];
vector<int> a[100015], s[100015];
void dfsLCA(int u) {
  num[u] = ++dfsCnt;
  cnt[u] = 1;
  s[h[u]].push_back(num[u]);
  for (int v : a[u]) h[v] = h[u] + 1, dfsLCA(v), cnt[u] += cnt[v];
}
void initLCA() {
  for (int u = 1; u <= n; ++u)
    if (!p[0][u]) h[u] = 1, dfsLCA(u);
  for (int k = 1, p2k = 2; p2k <= n; ++k, p2k <<= 1)
    for (int u = 1; u <= n; ++u) p[k][u] = p[k - 1][p[k - 1][u]];
}
int ancestor(int u, int x) {
  for (int k; x; x ^= 1 << k) k = log2(x), u = p[k][u];
  return u;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> p[0][i], a[p[0][i]].push_back(i);
  initLCA();
  for (int i = 1; i <= n; ++i) sort(s[i].begin(), s[i].end());
  cin >> m;
  for (int v, k, x; m--;) {
    cin >> v >> k;
    x = ancestor(v, k);
    if (!x)
      cout << "0 ";
    else
      cout << lower_bound(s[h[v]].begin(), s[h[v]].end(), num[x] + cnt[x]) -
                  lower_bound(s[h[v]].begin(), s[h[v]].end(), num[x]) - 1
           << ' ';
  }
}

```