#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100;
int col[N];
vector<int> G[N];
int sz[N], son[N], dep[N], fa[N][20];
void dfs(int x, int ff) {
  for (int i = 1; i <= 19; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  sz[x] = 1;
  for (int v : G[x]) {
    if (v == ff) continue;
    dep[v] = dep[x] + 1;
    fa[v][0] = x;
    dfs(v, x);
    sz[x] += sz[v];
    if (sz[son[x]] < sz[v]) son[x] = v;
  }
}
int Son;
int cnt[N];
int ans[N];
vector<pair<int, int>> Q[N];
void add(int x, int fa, int val) {
  cnt[dep[x]] += val;
  for (int v : G[x]) {
    if (v == fa || Son == v) continue;
    add(v, x, val);
  }
}
void DFS(int x, int fa, int key) {
  for (int v : G[x]) {
    if (v == fa || v == son[x]) continue;
    DFS(v, x, 0);
  }
  if (son[x]) DFS(son[x], x, 1), Son = son[x];
  add(x, fa, 1);
  Son = 0;
  for (auto p : Q[x]) {
    int d = p.first, id = p.second;
    ans[id] = cnt[d] - 1;
  }
  if (key == 0) add(x, fa, -1);
}
int rt[N];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int rtcnt = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x == 0)
      rt[++rtcnt] = i;
    else {
      G[x].push_back(i);
      G[i].push_back(x);
    }
  }
  for (int i = 1; i <= rtcnt; i++) dfs(rt[i], 0);
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    int d = dep[x];
    for (int i = 19; i >= 0; i--)
      if (y >> i & 1) x = fa[x][i];
    Q[x].push_back({d, i});
  }
  for (int i = 1; i <= rtcnt; i++) DFS(rt[i], 0, 0);
  for (int i = 1; i <= m; i++) cout << ans[i] << ' ';
  return 0;
}

### Prompt

Your task is to create 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 N = 1e5 + 100;
int col[N];
vector<int> G[N];
int sz[N], son[N], dep[N], fa[N][20];
void dfs(int x, int ff) {
  for (int i = 1; i <= 19; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  sz[x] = 1;
  for (int v : G[x]) {
    if (v == ff) continue;
    dep[v] = dep[x] + 1;
    fa[v][0] = x;
    dfs(v, x);
    sz[x] += sz[v];
    if (sz[son[x]] < sz[v]) son[x] = v;
  }
}
int Son;
int cnt[N];
int ans[N];
vector<pair<int, int>> Q[N];
void add(int x, int fa, int val) {
  cnt[dep[x]] += val;
  for (int v : G[x]) {
    if (v == fa || Son == v) continue;
    add(v, x, val);
  }
}
void DFS(int x, int fa, int key) {
  for (int v : G[x]) {
    if (v == fa || v == son[x]) continue;
    DFS(v, x, 0);
  }
  if (son[x]) DFS(son[x], x, 1), Son = son[x];
  add(x, fa, 1);
  Son = 0;
  for (auto p : Q[x]) {
    int d = p.first, id = p.second;
    ans[id] = cnt[d] - 1;
  }
  if (key == 0) add(x, fa, -1);
}
int rt[N];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int rtcnt = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x == 0)
      rt[++rtcnt] = i;
    else {
      G[x].push_back(i);
      G[i].push_back(x);
    }
  }
  for (int i = 1; i <= rtcnt; i++) dfs(rt[i], 0);
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    int d = dep[x];
    for (int i = 19; i >= 0; i--)
      if (y >> i & 1) x = fa[x][i];
    Q[x].push_back({d, i});
  }
  for (int i = 1; i <= rtcnt; i++) DFS(rt[i], 0, 0);
  for (int i = 1; i <= m; i++) cout << ans[i] << ' ';
  return 0;
}

```