#include <bits/stdc++.h>
using namespace std;
const int N = 100002;
int n, m, perTable[N][20], levelIndex[N], level[N];
vector<int> adjL[N], levelNodes[N], roots;
void dfs(int node, int depth) {
  level[node] = depth;
  levelIndex[node] = ((int)(levelNodes[depth]).size());
  levelNodes[depth].push_back(node);
  for (int ch : adjL[node]) dfs(ch, depth + 1);
}
void build() {
  for (int j = 1; (1 << j) <= n; ++j)
    for (int i = (1); i <= (int)(n); ++i)
      perTable[i][j] = perTable[perTable[i][j - 1]][j - 1];
}
int getKthPer(int node, int k) {
  for (int i = 0; (1 << i) <= k; ++i)
    if (k & (1 << i)) node = perTable[node][i];
  return node;
}
int getFarestNodeRight(int l, int r, int p, int lev, int x) {
  while (l < r - 1) {
    int m = l + ((r - l) >> 1);
    if (getKthPer(levelNodes[lev][m], p) != x)
      r = m - 1;
    else
      l = m;
  }
  if (getKthPer(levelNodes[lev][r], p) == x) l = r;
  return l;
}
int getFarestNodeLeft(int l, int r, int p, int lev, int x) {
  while (l < r) {
    int m = l + ((r - l) >> 1);
    if (getKthPer(levelNodes[lev][m], p) != x)
      l = m + 1;
    else
      r = m;
  }
  return l;
}
int solve(int node, int i, int l, int p) {
  int per = getKthPer(node, p);
  if (!per) return 0;
  int leftMost = getFarestNodeLeft(0, i, p, l, per);
  int rightMost =
      getFarestNodeRight(i, ((int)(levelNodes[l]).size()) - 1, p, l, per);
  return rightMost - leftMost;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (int)(n); ++i) {
    scanf("%d", &perTable[i][0]);
    if (perTable[i][0])
      adjL[perTable[i][0]].push_back(i);
    else
      roots.push_back(i);
  }
  build();
  for (int r : roots) dfs(r, 1);
  scanf("%d", &m);
  while (m--) {
    int v, p;
    scanf("%d%d", &v, &p);
    printf("%d ", solve(v, levelIndex[v], level[v], p));
  }
  printf("\n");
  return 0;
}

### 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;
const int N = 100002;
int n, m, perTable[N][20], levelIndex[N], level[N];
vector<int> adjL[N], levelNodes[N], roots;
void dfs(int node, int depth) {
  level[node] = depth;
  levelIndex[node] = ((int)(levelNodes[depth]).size());
  levelNodes[depth].push_back(node);
  for (int ch : adjL[node]) dfs(ch, depth + 1);
}
void build() {
  for (int j = 1; (1 << j) <= n; ++j)
    for (int i = (1); i <= (int)(n); ++i)
      perTable[i][j] = perTable[perTable[i][j - 1]][j - 1];
}
int getKthPer(int node, int k) {
  for (int i = 0; (1 << i) <= k; ++i)
    if (k & (1 << i)) node = perTable[node][i];
  return node;
}
int getFarestNodeRight(int l, int r, int p, int lev, int x) {
  while (l < r - 1) {
    int m = l + ((r - l) >> 1);
    if (getKthPer(levelNodes[lev][m], p) != x)
      r = m - 1;
    else
      l = m;
  }
  if (getKthPer(levelNodes[lev][r], p) == x) l = r;
  return l;
}
int getFarestNodeLeft(int l, int r, int p, int lev, int x) {
  while (l < r) {
    int m = l + ((r - l) >> 1);
    if (getKthPer(levelNodes[lev][m], p) != x)
      l = m + 1;
    else
      r = m;
  }
  return l;
}
int solve(int node, int i, int l, int p) {
  int per = getKthPer(node, p);
  if (!per) return 0;
  int leftMost = getFarestNodeLeft(0, i, p, l, per);
  int rightMost =
      getFarestNodeRight(i, ((int)(levelNodes[l]).size()) - 1, p, l, per);
  return rightMost - leftMost;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (int)(n); ++i) {
    scanf("%d", &perTable[i][0]);
    if (perTable[i][0])
      adjL[perTable[i][0]].push_back(i);
    else
      roots.push_back(i);
  }
  build();
  for (int r : roots) dfs(r, 1);
  scanf("%d", &m);
  while (m--) {
    int v, p;
    scanf("%d%d", &v, &p);
    printf("%d ", solve(v, levelIndex[v], level[v], p));
  }
  printf("\n");
  return 0;
}

```