#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 7;
const int maxl = 21;
int tin[MAXN], tout[MAXN], depth[MAXN], up[MAXN][maxl];
int timer = 1;
vector<int> v[MAXN], adj[MAXN];
void dfs(int u, int p) {
  tin[u] = timer++;
  depth[u] = depth[p] + 1;
  v[depth[u]].push_back(tin[u]);
  up[u][0] = p;
  for (int i = 1; i < maxl; i++) {
    up[u][i] = up[up[u][i - 1]][i - 1];
  }
  for (int v : adj[u]) {
    if (v != p) {
      dfs(v, u);
    }
  }
  tout[u] = timer++;
}
int get(int u, int p) {
  int curr = u;
  for (int i = maxl - 1; i >= 0; i--) {
    if ((1 << i) <= p) {
      p -= (1 << i);
      u = up[u][i];
    }
  }
  return u;
}
int query(int u, int pth) {
  int low = lower_bound(v[depth[u]].begin(), v[depth[u]].end(), tin[pth]) -
            v[depth[u]].begin();
  int up = upper_bound(v[depth[u]].begin(), v[depth[u]].end(), tout[pth]) -
           v[depth[u]].begin();
  return up - low - 1;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x != 0) {
      adj[x].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (tin[i] == 0) {
      dfs(i, 0);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int u, p;
    scanf("%d %d", &u, &p);
    if (depth[u] <= p) {
      cout << "0 ";
    } else {
      int pth = get(u, p);
      cout << query(u, pth) << ' ';
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 + 7;
const int maxl = 21;
int tin[MAXN], tout[MAXN], depth[MAXN], up[MAXN][maxl];
int timer = 1;
vector<int> v[MAXN], adj[MAXN];
void dfs(int u, int p) {
  tin[u] = timer++;
  depth[u] = depth[p] + 1;
  v[depth[u]].push_back(tin[u]);
  up[u][0] = p;
  for (int i = 1; i < maxl; i++) {
    up[u][i] = up[up[u][i - 1]][i - 1];
  }
  for (int v : adj[u]) {
    if (v != p) {
      dfs(v, u);
    }
  }
  tout[u] = timer++;
}
int get(int u, int p) {
  int curr = u;
  for (int i = maxl - 1; i >= 0; i--) {
    if ((1 << i) <= p) {
      p -= (1 << i);
      u = up[u][i];
    }
  }
  return u;
}
int query(int u, int pth) {
  int low = lower_bound(v[depth[u]].begin(), v[depth[u]].end(), tin[pth]) -
            v[depth[u]].begin();
  int up = upper_bound(v[depth[u]].begin(), v[depth[u]].end(), tout[pth]) -
           v[depth[u]].begin();
  return up - low - 1;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x != 0) {
      adj[x].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (tin[i] == 0) {
      dfs(i, 0);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int u, p;
    scanf("%d %d", &u, &p);
    if (depth[u] <= p) {
      cout << "0 ";
    } else {
      int pth = get(u, p);
      cout << query(u, pth) << ' ';
    }
  }
  return 0;
}

```