#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int n, m, a, b;
vector<int> g[MAXN];
int sz[MAXN], dp[MAXN];
int cnt[MAXN];
vector<int> *vec[MAXN];
int par[MAXN][17];
vector<pair<int, int>> q[MAXN];
int ans[MAXN];
void get_size_depth(int u, int f) {
  sz[u] = 1;
  dp[u] = dp[f] + 1;
  for (auto &v : g[u]) {
    if (v == f) continue;
    get_size_depth(v, u);
    sz[u] += sz[v];
  }
}
void dfs(int u, int f, bool keep) {
  int mx = -1, big = -1;
  for (auto &v : g[u]) {
    if (v == f) continue;
    if (mx < sz[v]) big = v, mx = sz[v];
  }
  for (auto &v : g[u]) {
    if (v == f || v == big) continue;
    dfs(v, u, 0);
  }
  if (big != -1)
    dfs(big, u, 1), vec[u] = vec[big];
  else
    vec[u] = new vector<int>();
  vec[u]->push_back(u);
  cnt[dp[u]]++;
  for (auto &v : g[u]) {
    if (v == f || v == big) continue;
    for (auto &x : *vec[v]) {
      vec[u]->push_back(x);
      cnt[dp[x]]++;
    }
  }
  for (auto &p : q[u])
    ans[p.second] = (cnt[p.first + dp[u]] > 0) ? cnt[p.first + dp[u]] - 1 : 0;
  if (!keep)
    for (auto &x : *vec[u]) cnt[dp[x]]--;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    g[a].push_back(i);
    g[i].push_back(a);
    par[i][0] = a;
  }
  for (auto &p : g[0]) get_size_depth(p, 0);
  for (int u = 1; u <= n; u++)
    for (int i = 1; i < 17; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    int vh = dp[a];
    for (int j = 0; j < 17; j++)
      if ((vh - b - dp[a]) >> j & 1) a = par[a][j];
    q[a].push_back(make_pair(b, i));
  }
  for (auto &p : g[0]) dfs(p, 0, 0);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
}

### Prompt

Generate 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 MAXN = 1e5 + 10;
int n, m, a, b;
vector<int> g[MAXN];
int sz[MAXN], dp[MAXN];
int cnt[MAXN];
vector<int> *vec[MAXN];
int par[MAXN][17];
vector<pair<int, int>> q[MAXN];
int ans[MAXN];
void get_size_depth(int u, int f) {
  sz[u] = 1;
  dp[u] = dp[f] + 1;
  for (auto &v : g[u]) {
    if (v == f) continue;
    get_size_depth(v, u);
    sz[u] += sz[v];
  }
}
void dfs(int u, int f, bool keep) {
  int mx = -1, big = -1;
  for (auto &v : g[u]) {
    if (v == f) continue;
    if (mx < sz[v]) big = v, mx = sz[v];
  }
  for (auto &v : g[u]) {
    if (v == f || v == big) continue;
    dfs(v, u, 0);
  }
  if (big != -1)
    dfs(big, u, 1), vec[u] = vec[big];
  else
    vec[u] = new vector<int>();
  vec[u]->push_back(u);
  cnt[dp[u]]++;
  for (auto &v : g[u]) {
    if (v == f || v == big) continue;
    for (auto &x : *vec[v]) {
      vec[u]->push_back(x);
      cnt[dp[x]]++;
    }
  }
  for (auto &p : q[u])
    ans[p.second] = (cnt[p.first + dp[u]] > 0) ? cnt[p.first + dp[u]] - 1 : 0;
  if (!keep)
    for (auto &x : *vec[u]) cnt[dp[x]]--;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    g[a].push_back(i);
    g[i].push_back(a);
    par[i][0] = a;
  }
  for (auto &p : g[0]) get_size_depth(p, 0);
  for (int u = 1; u <= n; u++)
    for (int i = 1; i < 17; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    int vh = dp[a];
    for (int j = 0; j < 17; j++)
      if ((vh - b - dp[a]) >> j & 1) a = par[a][j];
    q[a].push_back(make_pair(b, i));
  }
  for (auto &p : g[0]) dfs(p, 0, 0);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
}

```