#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
int n, m, root;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) (ch == '-') && (f = -f), ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int head[MAXN], trump;
struct Edge {
  int to, next;
  Edge(int u = 0, int v = 0) : to(v), next(head[u]){};
} e[MAXN];
inline void addedge(int u, int v) {
  u = u ? u : root;
  e[++trump] = Edge(u, v);
  head[u] = trump;
}
int prt[MAXN];
struct Query {
  int k, ord;
  Query(int k = 0, int ord = 0) : k(k), ord(ord){};
};
vector<Query> S[MAXN];
vector<Query> Q[MAXN];
int dep[MAXN], size[MAXN], son[MAXN];
int s[MAXN], top;
void tarjan(int u) {
  s[++top] = u;
  size[u] = 1;
  for (vector<Query>::iterator it = S[u].begin(); it != S[u].end(); ++it) {
    int k = (*it).k, ord = (*it).ord;
    if (top - k > 1) Q[s[top - k]].push_back(Query(k, ord));
  }
  for (register int i = head[u]; i; i = e[i].next) {
    dep[e[i].to] = dep[u] + 1;
    tarjan(e[i].to);
    size[u] += size[e[i].to];
    (size[son[u]] < size[e[i].to]) && (son[u] = e[i].to);
  }
  s[top--] = 0;
}
int cnt[MAXN];
void work(int u, int w) {
  ++cnt[dep[u]];
  for (register int i = head[u]; i; i = e[i].next)
    (e[i].to != w) && (work(e[i].to, w), 1);
}
void clear(int u) {
  --cnt[dep[u]];
  for (register int i = head[u]; i; i = e[i].next) clear(e[i].to);
}
void dsu(int u) {
  for (register int i = head[u]; i; i = e[i].next)
    (e[i].to != son[u]) && (dsu(e[i].to), clear(e[i].to), 1);
  son[u] && (dsu(son[u]), 1);
  work(u, son[u]);
  for (vector<Query>::iterator it = Q[u].begin(); it != Q[u].end(); ++it)
    prt[(*it).ord] = cnt[dep[u] + (*it).k] - 1;
}
int main() {
  register int i;
  n = read();
  root = n + 1;
  for (i = 1; i <= n; ++i) addedge(read(), i);
  m = read();
  for (i = 1; i <= m; ++i) {
    int u = read(), k = read();
    S[u].push_back(Query(k, i));
  }
  tarjan(root);
  dsu(root);
  for (i = 1; i <= m; ++i) printf("%d ", prt[i]);
  return 0;
}

### 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;
const int MAXN = 100005;
int n, m, root;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) (ch == '-') && (f = -f), ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int head[MAXN], trump;
struct Edge {
  int to, next;
  Edge(int u = 0, int v = 0) : to(v), next(head[u]){};
} e[MAXN];
inline void addedge(int u, int v) {
  u = u ? u : root;
  e[++trump] = Edge(u, v);
  head[u] = trump;
}
int prt[MAXN];
struct Query {
  int k, ord;
  Query(int k = 0, int ord = 0) : k(k), ord(ord){};
};
vector<Query> S[MAXN];
vector<Query> Q[MAXN];
int dep[MAXN], size[MAXN], son[MAXN];
int s[MAXN], top;
void tarjan(int u) {
  s[++top] = u;
  size[u] = 1;
  for (vector<Query>::iterator it = S[u].begin(); it != S[u].end(); ++it) {
    int k = (*it).k, ord = (*it).ord;
    if (top - k > 1) Q[s[top - k]].push_back(Query(k, ord));
  }
  for (register int i = head[u]; i; i = e[i].next) {
    dep[e[i].to] = dep[u] + 1;
    tarjan(e[i].to);
    size[u] += size[e[i].to];
    (size[son[u]] < size[e[i].to]) && (son[u] = e[i].to);
  }
  s[top--] = 0;
}
int cnt[MAXN];
void work(int u, int w) {
  ++cnt[dep[u]];
  for (register int i = head[u]; i; i = e[i].next)
    (e[i].to != w) && (work(e[i].to, w), 1);
}
void clear(int u) {
  --cnt[dep[u]];
  for (register int i = head[u]; i; i = e[i].next) clear(e[i].to);
}
void dsu(int u) {
  for (register int i = head[u]; i; i = e[i].next)
    (e[i].to != son[u]) && (dsu(e[i].to), clear(e[i].to), 1);
  son[u] && (dsu(son[u]), 1);
  work(u, son[u]);
  for (vector<Query>::iterator it = Q[u].begin(); it != Q[u].end(); ++it)
    prt[(*it).ord] = cnt[dep[u] + (*it).k] - 1;
}
int main() {
  register int i;
  n = read();
  root = n + 1;
  for (i = 1; i <= n; ++i) addedge(read(), i);
  m = read();
  for (i = 1; i <= m; ++i) {
    int u = read(), k = read();
    S[u].push_back(Query(k, i));
  }
  tarjan(root);
  dsu(root);
  for (i = 1; i <= m; ++i) printf("%d ", prt[i]);
  return 0;
}

```