#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
vector<int> g[N];
struct query {
  int dep, id;
};
vector<query> q[N];
int son[N], sz[N], ans[N], deep[N], father[N], n, fn[N][30];
void dfs(int u, int fa) {
  sz[u] = 1;
  deep[u] = deep[fa] + 1;
  fn[u][0] = fa;
  for (int i = 1; i < 30; i++) {
    fn[u][i] = fn[fn[u][i - 1]][i - 1];
  }
  for (int to : g[u]) {
    if (to == fa) continue;
    dfs(to, u);
    sz[u] += sz[to];
    if (sz[son[u]] < sz[to]) {
      son[u] = to;
    }
  }
}
int cnt[N];
void work(int v, int u, int fa, int zson) {
  cnt[deep[u]] += v;
  for (int to : g[u]) {
    if (to == fa || to == zson) continue;
    work(v, to, u, zson);
  }
}
void gao(int u) {
  for (int i = 0; i < q[u].size(); i++) {
    int id = q[u][i].id;
    int dep = deep[q[u][i].dep];
    ans[id] = cnt[dep] - 1;
  }
}
void dfs1(int u, int fa, int ok) {
  for (int to : g[u]) {
    if (to == fa || to == son[u]) {
      continue;
    }
    dfs1(to, u, 1);
  }
  if (son[u]) {
    dfs1(son[u], u, 0);
  }
  work(1, u, fa, son[u]);
  gao(u);
  if (ok) {
    work(-1, u, fa, -1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int u;
    scanf("%d", &u);
    father[i] = u;
    if (u) {
      g[i].push_back(u);
      g[u].push_back(i);
    }
  }
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= n; i++) {
    if (father[i] == 0) {
      dfs(i, 0);
    }
  }
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    int now = u;
    int cn = 0;
    while (v) {
      if (v & 1) {
        now = fn[now][cn];
      }
      cn++;
      v = v / 2;
    }
    q[now].push_back({u, i});
  }
  for (int i = 1; i <= n; i++) {
    if (father[i] == 0) {
      dfs1(i, 0, 1);
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d ", ans[i]);
  }
  puts("");
}

### Prompt

Your challenge is to write 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 + 7;
vector<int> g[N];
struct query {
  int dep, id;
};
vector<query> q[N];
int son[N], sz[N], ans[N], deep[N], father[N], n, fn[N][30];
void dfs(int u, int fa) {
  sz[u] = 1;
  deep[u] = deep[fa] + 1;
  fn[u][0] = fa;
  for (int i = 1; i < 30; i++) {
    fn[u][i] = fn[fn[u][i - 1]][i - 1];
  }
  for (int to : g[u]) {
    if (to == fa) continue;
    dfs(to, u);
    sz[u] += sz[to];
    if (sz[son[u]] < sz[to]) {
      son[u] = to;
    }
  }
}
int cnt[N];
void work(int v, int u, int fa, int zson) {
  cnt[deep[u]] += v;
  for (int to : g[u]) {
    if (to == fa || to == zson) continue;
    work(v, to, u, zson);
  }
}
void gao(int u) {
  for (int i = 0; i < q[u].size(); i++) {
    int id = q[u][i].id;
    int dep = deep[q[u][i].dep];
    ans[id] = cnt[dep] - 1;
  }
}
void dfs1(int u, int fa, int ok) {
  for (int to : g[u]) {
    if (to == fa || to == son[u]) {
      continue;
    }
    dfs1(to, u, 1);
  }
  if (son[u]) {
    dfs1(son[u], u, 0);
  }
  work(1, u, fa, son[u]);
  gao(u);
  if (ok) {
    work(-1, u, fa, -1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int u;
    scanf("%d", &u);
    father[i] = u;
    if (u) {
      g[i].push_back(u);
      g[u].push_back(i);
    }
  }
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= n; i++) {
    if (father[i] == 0) {
      dfs(i, 0);
    }
  }
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    int now = u;
    int cn = 0;
    while (v) {
      if (v & 1) {
        now = fn[now][cn];
      }
      cn++;
      v = v / 2;
    }
    q[now].push_back({u, i});
  }
  for (int i = 1; i <= n; i++) {
    if (father[i] == 0) {
      dfs1(i, 0, 1);
    }
  }
  for (int i = 1; i <= m; i++) {
    printf("%d ", ans[i]);
  }
  puts("");
}

```