#include <bits/stdc++.h>
using namespace std;
int n, x, pr[21][111111], de[111111], t[111111], le[111111], ri[111111], tt, m,
    y;
vector<int> g[111111], v[111111];
void rec(int x, int d) {
  t[x] = ++tt;
  le[x] = tt;
  de[x] = d;
  for (int i = 0; i < v[x].size(); i++) rec(v[x][i], d + 1);
  ri[x] = tt;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    pr[0][i] = x;
    v[x].push_back(i);
  }
  rec(0, 0);
  for (int i = 1; i < 20; i++)
    for (int j = 0; j <= n; j++)
      if (!pr[i - 1][j])
        pr[i][j] = 0;
      else
        pr[i][j] = pr[i - 1][pr[i - 1][j]];
  for (int i = 0; i <= n; i++) g[de[i]].push_back(t[i]);
  for (int i = 0; i <= n + 2; i++)
    if (g[i].size()) sort(g[i].begin(), g[i].end());
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    if (de[x] <= y) {
      printf("0 ");
      continue;
    }
    int xx = x;
    for (int j = 19; j >= 0; j--)
      if (y & (1ull << (j))) xx = pr[j][xx];
    int ans = upper_bound(g[de[x]].begin(), g[de[x]].end(), ri[xx]) -
              lower_bound(g[de[x]].begin(), g[de[x]].end(), le[xx]);
    printf("%d ", ans - 1);
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 n, x, pr[21][111111], de[111111], t[111111], le[111111], ri[111111], tt, m,
    y;
vector<int> g[111111], v[111111];
void rec(int x, int d) {
  t[x] = ++tt;
  le[x] = tt;
  de[x] = d;
  for (int i = 0; i < v[x].size(); i++) rec(v[x][i], d + 1);
  ri[x] = tt;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    pr[0][i] = x;
    v[x].push_back(i);
  }
  rec(0, 0);
  for (int i = 1; i < 20; i++)
    for (int j = 0; j <= n; j++)
      if (!pr[i - 1][j])
        pr[i][j] = 0;
      else
        pr[i][j] = pr[i - 1][pr[i - 1][j]];
  for (int i = 0; i <= n; i++) g[de[i]].push_back(t[i]);
  for (int i = 0; i <= n + 2; i++)
    if (g[i].size()) sort(g[i].begin(), g[i].end());
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    if (de[x] <= y) {
      printf("0 ");
      continue;
    }
    int xx = x;
    for (int j = 19; j >= 0; j--)
      if (y & (1ull << (j))) xx = pr[j][xx];
    int ans = upper_bound(g[de[x]].begin(), g[de[x]].end(), ri[xx]) -
              lower_bound(g[de[x]].begin(), g[de[x]].end(), le[xx]);
    printf("%d ", ans - 1);
  }
  return 0;
}

```