#include <bits/stdc++.h>
using namespace std;
vector<int> v[100005];
vector<int> dep_t[100005];
int dep[100005], p[100005][20];
int in[100005], out[100005];
int ans[100005];
vector<int> roots;
int times;
void dfs(int u, int fa, int d) {
  in[u] = ++times;
  p[u][0] = fa;
  for (int i = 1; i < 20; i++) {
    p[u][i] = p[p[u][i - 1]][i - 1];
  }
  dep[u] = d;
  dep_t[d].push_back(in[u]);
  for (int i = 0; i < v[u].size(); i++) {
    dfs(v[u][i], u, d + 1);
  }
  out[u] = ++times;
}
int find(int v, int d) {
  for (int i = 0; i < 20; i++) {
    if ((d >> i) & 1) {
      v = p[v][i];
    }
  }
  return v;
}
int cal(int v, int d) {
  int l = 0, r = dep_t[d].size() - 1;
  if (v < dep_t[d][0]) return 0;
  while (l < r) {
    int m = (l + r + 1) >> 1;
    if (dep_t[d][m] <= v) {
      l = m;
    } else {
      r = m - 1;
    }
  }
  return l + 1;
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) {
      v[i].clear();
      dep_t[i].clear();
    }
    roots.clear();
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
    memset(dep, 0, sizeof(dep));
    memset(p, 0, sizeof(p));
    for (int i = 1; i <= n; i++) {
      int u;
      scanf("%d", &u);
      if (u == 0) {
        roots.push_back(i);
      } else {
        v[u].push_back(i);
      }
    }
    times = 0;
    for (int i = 0; i < roots.size(); i++) {
      dfs(roots[i], 0, 1);
    }
    int m;
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
      int v, pp;
      scanf("%d%d", &v, &pp);
      int vv = find(v, pp);
      if (vv == 0) {
        ans[i] = 0;
      } else {
        ans[i] = cal(out[vv], dep[v]) - cal(in[vv], dep[v]) - 1;
      }
    }
    for (int i = 0; i < m; i++) {
      printf(i == m - 1 ? "%d\n" : "%d ", ans[i]);
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
vector<int> v[100005];
vector<int> dep_t[100005];
int dep[100005], p[100005][20];
int in[100005], out[100005];
int ans[100005];
vector<int> roots;
int times;
void dfs(int u, int fa, int d) {
  in[u] = ++times;
  p[u][0] = fa;
  for (int i = 1; i < 20; i++) {
    p[u][i] = p[p[u][i - 1]][i - 1];
  }
  dep[u] = d;
  dep_t[d].push_back(in[u]);
  for (int i = 0; i < v[u].size(); i++) {
    dfs(v[u][i], u, d + 1);
  }
  out[u] = ++times;
}
int find(int v, int d) {
  for (int i = 0; i < 20; i++) {
    if ((d >> i) & 1) {
      v = p[v][i];
    }
  }
  return v;
}
int cal(int v, int d) {
  int l = 0, r = dep_t[d].size() - 1;
  if (v < dep_t[d][0]) return 0;
  while (l < r) {
    int m = (l + r + 1) >> 1;
    if (dep_t[d][m] <= v) {
      l = m;
    } else {
      r = m - 1;
    }
  }
  return l + 1;
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) {
      v[i].clear();
      dep_t[i].clear();
    }
    roots.clear();
    memset(in, 0, sizeof(in));
    memset(out, 0, sizeof(out));
    memset(dep, 0, sizeof(dep));
    memset(p, 0, sizeof(p));
    for (int i = 1; i <= n; i++) {
      int u;
      scanf("%d", &u);
      if (u == 0) {
        roots.push_back(i);
      } else {
        v[u].push_back(i);
      }
    }
    times = 0;
    for (int i = 0; i < roots.size(); i++) {
      dfs(roots[i], 0, 1);
    }
    int m;
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
      int v, pp;
      scanf("%d%d", &v, &pp);
      int vv = find(v, pp);
      if (vv == 0) {
        ans[i] = 0;
      } else {
        ans[i] = cal(out[vv], dep[v]) - cal(in[vv], dep[v]) - 1;
      }
    }
    for (int i = 0; i < m; i++) {
      printf(i == m - 1 ? "%d\n" : "%d ", ans[i]);
    }
  }
  return 0;
}

```