#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const int maxh = 17;
const int inf = 1 << 29;
const int mod = 1000000007;
const double eps = 1e-8;
vector<int> v[maxn];
int par[maxn][maxh], dep[maxn], n, fd[maxn], fin[maxn], ti;
vector<int> h[maxn];
void dfs(int x, int p, int d) {
  par[x][0] = p;
  dep[x] = d;
  fd[x] = ++ti;
  h[d].push_back(ti);
  for (int i = 0; i < v[x].size(); i++) dfs(v[x][i], x, d + 1);
  fin[x] = ti;
}
void pre() {
  for (int k = 0; k + 1 < maxh; k++)
    for (int i = 1; i <= n; i++)
      par[i][k + 1] = par[i][k] <= 0 ? 0 : par[par[i][k]][k];
}
int query(int a, int b) {
  int d = dep[a];
  if (d <= b) return 0;
  for (int i = 0; i < maxh; i++)
    if (b & (1 << i)) a = par[a][i];
  return upper_bound(h[d].begin(), h[d].end(), fin[a]) -
         lower_bound(h[d].begin(), h[d].end(), fd[a]) - 1;
}
int main() {
  int q, a, b;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    v[a].push_back(i);
  }
  for (int i = 0; i < v[0].size(); i++) dfs(v[0][i], 0, 1);
  pre();
  for (int i = 1; i <= n; i++) sort(h[i].begin(), h[i].end());
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &a, &b);
    printf("%d ", query(a, b));
  }
}

### 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 = 100005;
const int maxh = 17;
const int inf = 1 << 29;
const int mod = 1000000007;
const double eps = 1e-8;
vector<int> v[maxn];
int par[maxn][maxh], dep[maxn], n, fd[maxn], fin[maxn], ti;
vector<int> h[maxn];
void dfs(int x, int p, int d) {
  par[x][0] = p;
  dep[x] = d;
  fd[x] = ++ti;
  h[d].push_back(ti);
  for (int i = 0; i < v[x].size(); i++) dfs(v[x][i], x, d + 1);
  fin[x] = ti;
}
void pre() {
  for (int k = 0; k + 1 < maxh; k++)
    for (int i = 1; i <= n; i++)
      par[i][k + 1] = par[i][k] <= 0 ? 0 : par[par[i][k]][k];
}
int query(int a, int b) {
  int d = dep[a];
  if (d <= b) return 0;
  for (int i = 0; i < maxh; i++)
    if (b & (1 << i)) a = par[a][i];
  return upper_bound(h[d].begin(), h[d].end(), fin[a]) -
         lower_bound(h[d].begin(), h[d].end(), fd[a]) - 1;
}
int main() {
  int q, a, b;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    v[a].push_back(i);
  }
  for (int i = 0; i < v[0].size(); i++) dfs(v[0][i], 0, 1);
  pre();
  for (int i = 1; i <= n; i++) sort(h[i].begin(), h[i].end());
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &a, &b);
    printf("%d ", query(a, b));
  }
}

```