#include <bits/stdc++.h>
using namespace std;
int tr[20][100005];
vector<int> way[100005];
vector<int> h[100005];
int hinf[100005];
int rnk[100005];
void dfs(int pos, int from, int cur) {
  h[cur].push_back(pos);
  hinf[pos] = cur;
  rnk[pos] = (int)h[cur].size();
  for (int i = 0; i < way[pos].size(); i++) dfs(way[pos][i], pos, cur + 1);
  return;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &tr[0][i]);
    way[tr[0][i]].push_back(i);
  }
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++) tr[i][j] = tr[i - 1][tr[i - 1][j]];
  for (int i = 1; i <= n; i++)
    if (!tr[0][i]) dfs(i, 0, 0);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int v, p;
    scanf("%d%d", &v, &p);
    int anc = v;
    int temp = p;
    int cur = 0;
    while (temp) {
      if (temp & 1) anc = tr[cur][anc];
      cur++;
      temp >>= 1;
    }
    if (anc == 0) {
      printf("0 ");
      continue;
    }
    int l = 0, r = (int)h[p + hinf[anc]].size() - 1;
    int rbound = 1000000, lbound = -1;
    while (l <= r) {
      int m = (l + r) / 2;
      int curpos = h[p + hinf[anc]][m];
      int temp = p;
      int cur = 0;
      while (temp) {
        if (temp & 1) curpos = tr[cur][curpos];
        cur++;
        temp >>= 1;
      }
      if (rnk[curpos] > rnk[anc])
        r = m - 1;
      else if (rnk[curpos] < rnk[anc])
        l = m + 1;
      else
        l = m + 1, lbound = m;
    }
    l = 0, r = (int)h[p + hinf[anc]].size() - 1;
    while (l <= r) {
      int m = (l + r) / 2;
      int curpos = h[p + hinf[anc]][m];
      int temp = p;
      int cur = 0;
      while (temp) {
        if (temp & 1) curpos = tr[cur][curpos];
        cur++;
        temp >>= 1;
      }
      if (rnk[curpos] > rnk[anc])
        r = m - 1;
      else if (rnk[curpos] < rnk[anc])
        l = m + 1;
      else
        r = m - 1, rbound = m;
    }
    printf("%d ", lbound - rbound);
  }
  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;
int tr[20][100005];
vector<int> way[100005];
vector<int> h[100005];
int hinf[100005];
int rnk[100005];
void dfs(int pos, int from, int cur) {
  h[cur].push_back(pos);
  hinf[pos] = cur;
  rnk[pos] = (int)h[cur].size();
  for (int i = 0; i < way[pos].size(); i++) dfs(way[pos][i], pos, cur + 1);
  return;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &tr[0][i]);
    way[tr[0][i]].push_back(i);
  }
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++) tr[i][j] = tr[i - 1][tr[i - 1][j]];
  for (int i = 1; i <= n; i++)
    if (!tr[0][i]) dfs(i, 0, 0);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int v, p;
    scanf("%d%d", &v, &p);
    int anc = v;
    int temp = p;
    int cur = 0;
    while (temp) {
      if (temp & 1) anc = tr[cur][anc];
      cur++;
      temp >>= 1;
    }
    if (anc == 0) {
      printf("0 ");
      continue;
    }
    int l = 0, r = (int)h[p + hinf[anc]].size() - 1;
    int rbound = 1000000, lbound = -1;
    while (l <= r) {
      int m = (l + r) / 2;
      int curpos = h[p + hinf[anc]][m];
      int temp = p;
      int cur = 0;
      while (temp) {
        if (temp & 1) curpos = tr[cur][curpos];
        cur++;
        temp >>= 1;
      }
      if (rnk[curpos] > rnk[anc])
        r = m - 1;
      else if (rnk[curpos] < rnk[anc])
        l = m + 1;
      else
        l = m + 1, lbound = m;
    }
    l = 0, r = (int)h[p + hinf[anc]].size() - 1;
    while (l <= r) {
      int m = (l + r) / 2;
      int curpos = h[p + hinf[anc]][m];
      int temp = p;
      int cur = 0;
      while (temp) {
        if (temp & 1) curpos = tr[cur][curpos];
        cur++;
        temp >>= 1;
      }
      if (rnk[curpos] > rnk[anc])
        r = m - 1;
      else if (rnk[curpos] < rnk[anc])
        l = m + 1;
      else
        r = m - 1, rbound = m;
    }
    printf("%d ", lbound - rbound);
  }
  return 0;
}

```