#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
vector<int> rt;
vector<int> dep[100050];
vector<int> g[100050];
int in[100050], out[100050], dfn = 0, p[100050][20], deep[100050];
void dfs(int x, int fa) {
  in[x] = ++dfn;
  p[x][0] = fa;
  deep[x] = deep[fa] + 1;
  dep[deep[x]].push_back(dfn);
  for (int j = 1; j <= 18; j++) p[x][j] = p[p[x][j - 1]][j - 1];
  for (int i = 0; i < g[x].size(); i++) {
    int v = g[x][i];
    if (v == fa) continue;
    dfs(v, x);
  }
  out[x] = dfn;
}
inline int finds(int x, int k) {
  for (int i = 18; i >= 0; i--)
    if (k & (1 << i)) x = p[x][i];
  return x;
}
inline int solve(int dfn, int d) {
  int l = 0, r = dep[d].size() - 1, ans = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (dep[d][mid] <= dfn)
      l = mid + 1, ans = mid;
    else
      r = mid - 1;
  }
  return ans;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    int x = read();
    if (x != 0)
      g[x].push_back(i);
    else
      rt.push_back(i);
  }
  for (int i = 0; i < rt.size(); i++) dfs(rt[i], 0);
  int m = read();
  while (m--) {
    int x = read(), p = read();
    int anc = finds(x, p);
    if (anc == 0)
      printf("0 ");
    else
      printf("%d ", solve(out[anc], deep[x]) - solve(in[anc], deep[x]) - 1);
  }
}

### Prompt

Generate a CPP solution 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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
vector<int> rt;
vector<int> dep[100050];
vector<int> g[100050];
int in[100050], out[100050], dfn = 0, p[100050][20], deep[100050];
void dfs(int x, int fa) {
  in[x] = ++dfn;
  p[x][0] = fa;
  deep[x] = deep[fa] + 1;
  dep[deep[x]].push_back(dfn);
  for (int j = 1; j <= 18; j++) p[x][j] = p[p[x][j - 1]][j - 1];
  for (int i = 0; i < g[x].size(); i++) {
    int v = g[x][i];
    if (v == fa) continue;
    dfs(v, x);
  }
  out[x] = dfn;
}
inline int finds(int x, int k) {
  for (int i = 18; i >= 0; i--)
    if (k & (1 << i)) x = p[x][i];
  return x;
}
inline int solve(int dfn, int d) {
  int l = 0, r = dep[d].size() - 1, ans = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (dep[d][mid] <= dfn)
      l = mid + 1, ans = mid;
    else
      r = mid - 1;
  }
  return ans;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    int x = read();
    if (x != 0)
      g[x].push_back(i);
    else
      rt.push_back(i);
  }
  for (int i = 0; i < rt.size(); i++) dfs(rt[i], 0);
  int m = read();
  while (m--) {
    int x = read(), p = read();
    int anc = finds(x, p);
    if (anc == 0)
      printf("0 ");
    else
      printf("%d ", solve(out[anc], deep[x]) - solve(in[anc], deep[x]) - 1);
  }
}

```