#include <bits/stdc++.h>
using namespace std;
int d[100100], cnt = 0, lv[100100], last[100100];
vector<int> b[100100];
vector<int> a[100100];
int par[100100][19];
void dfs(int x) {
  int i;
  d[x] = ++cnt;
  b[lv[x]].push_back(d[x]);
  for (i = 0; i < a[x].size(); i++) {
    lv[a[x][i]] = lv[x] + 1;
    par[a[x][i]][0] = x;
    dfs(a[x][i]);
  }
  last[x] = cnt;
}
int cal(int x, int y) {
  vector<int>::iterator itl = lower_bound(b[x].begin(), b[x].end(), d[y]);
  vector<int>::iterator itr = upper_bound(b[x].begin(), b[x].end(), last[y]);
  if (itl == b[x].end()) return 0;
  itr--;
  if (*itl > *itr) return 0;
  return itr - itl + 1;
}
int main() {
  int n, i, x, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[x].push_back(i);
  }
  lv[0] = 1;
  dfs(0);
  for (i = 1; i <= 17; i++)
    for (j = 0; j <= n; j++) {
      par[j][i] = par[par[j][i - 1]][i - 1];
    }
  int p, v, q;
  scanf("%d", &p);
  while (p--) {
    scanf("%d %d", &v, &q);
    q--;
    int tmp = lv[v];
    for (i = 17; i >= 0; i--) {
      if (q >= (1 << i)) {
        v = par[v][i];
        q -= (1 << i);
      }
    }
    if (par[v][0] == 0) {
      printf("0 ");
      continue;
    } else
      printf("%d ", cal(tmp, par[v][0]) - 1);
  }
}

### Prompt

In cpp, your task is to solve 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;
int d[100100], cnt = 0, lv[100100], last[100100];
vector<int> b[100100];
vector<int> a[100100];
int par[100100][19];
void dfs(int x) {
  int i;
  d[x] = ++cnt;
  b[lv[x]].push_back(d[x]);
  for (i = 0; i < a[x].size(); i++) {
    lv[a[x][i]] = lv[x] + 1;
    par[a[x][i]][0] = x;
    dfs(a[x][i]);
  }
  last[x] = cnt;
}
int cal(int x, int y) {
  vector<int>::iterator itl = lower_bound(b[x].begin(), b[x].end(), d[y]);
  vector<int>::iterator itr = upper_bound(b[x].begin(), b[x].end(), last[y]);
  if (itl == b[x].end()) return 0;
  itr--;
  if (*itl > *itr) return 0;
  return itr - itl + 1;
}
int main() {
  int n, i, x, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[x].push_back(i);
  }
  lv[0] = 1;
  dfs(0);
  for (i = 1; i <= 17; i++)
    for (j = 0; j <= n; j++) {
      par[j][i] = par[par[j][i - 1]][i - 1];
    }
  int p, v, q;
  scanf("%d", &p);
  while (p--) {
    scanf("%d %d", &v, &q);
    q--;
    int tmp = lv[v];
    for (i = 17; i >= 0; i--) {
      if (q >= (1 << i)) {
        v = par[v][i];
        q -= (1 << i);
      }
    }
    if (par[v][0] == 0) {
      printf("0 ");
      continue;
    } else
      printf("%d ", cal(tmp, par[v][0]) - 1);
  }
}

```