#include <bits/stdc++.h>
const int maxn = 1e5 + 5;
int read() {
  int s = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) x |= (c == '-'), c = getchar();
  while (isdigit(c)) s = s * 10 + c - '0', c = getchar();
  return x ? -s : s;
}
struct node {
  int id, k;
};
std::vector<node> query[maxn];
int n, fa[maxn];
int head[maxn], to[maxn], nxt[maxn], cnt;
int h[maxn], son[maxn], ans[maxn];
int *f[maxn], buf[maxn << 1], *now = buf;
inline void add(int u, int v) {
  to[++cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt;
  return;
}
int stk[maxn], top;
void dfs1(int u) {
  stk[++top] = u;
  for (int i = 0; i <= (int)query[u].size() - 1; ++i)
    if (query[u][i].k < top && query[u][i].k)
      query[stk[top - query[u][i].k]].push_back(query[u][i]);
  query[u].resize(0);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    dfs1(v);
    if (h[v] > h[son[u]]) son[u] = v;
  }
  h[u] = h[son[u]] + 1;
  --top;
  return;
}
void dfs2(int u) {
  f[u][0] = 1;
  if (son[u]) f[son[u]] = f[u] + 1, dfs2(son[u]);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == son[u]) continue;
    f[v] = now, now += h[v];
    dfs2(v);
    for (int i = 1; i <= h[v]; ++i) f[u][i] += f[v][i - 1];
  }
  for (int i = 0; i <= (int)query[u].size() - 1; ++i)
    ans[query[u][i].id] = f[u][query[u][i].k] - 1;
  return;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    fa[i] = read();
    if (fa[i]) add(fa[i], i);
  }
  int q = read();
  for (int i = 1; i <= q; ++i) {
    int u = read(), k = read();
    query[u].push_back((node){i, k});
  }
  for (int u = 1; u <= n; ++u)
    if (!fa[u]) dfs1(u), f[u] = now, now += h[u], dfs2(u);
  for (int i = 1; i <= q; ++i) printf("%d ", ans[i]);
  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>
const int maxn = 1e5 + 5;
int read() {
  int s = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) x |= (c == '-'), c = getchar();
  while (isdigit(c)) s = s * 10 + c - '0', c = getchar();
  return x ? -s : s;
}
struct node {
  int id, k;
};
std::vector<node> query[maxn];
int n, fa[maxn];
int head[maxn], to[maxn], nxt[maxn], cnt;
int h[maxn], son[maxn], ans[maxn];
int *f[maxn], buf[maxn << 1], *now = buf;
inline void add(int u, int v) {
  to[++cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt;
  return;
}
int stk[maxn], top;
void dfs1(int u) {
  stk[++top] = u;
  for (int i = 0; i <= (int)query[u].size() - 1; ++i)
    if (query[u][i].k < top && query[u][i].k)
      query[stk[top - query[u][i].k]].push_back(query[u][i]);
  query[u].resize(0);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    dfs1(v);
    if (h[v] > h[son[u]]) son[u] = v;
  }
  h[u] = h[son[u]] + 1;
  --top;
  return;
}
void dfs2(int u) {
  f[u][0] = 1;
  if (son[u]) f[son[u]] = f[u] + 1, dfs2(son[u]);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == son[u]) continue;
    f[v] = now, now += h[v];
    dfs2(v);
    for (int i = 1; i <= h[v]; ++i) f[u][i] += f[v][i - 1];
  }
  for (int i = 0; i <= (int)query[u].size() - 1; ++i)
    ans[query[u][i].id] = f[u][query[u][i].k] - 1;
  return;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    fa[i] = read();
    if (fa[i]) add(fa[i], i);
  }
  int q = read();
  for (int i = 1; i <= q; ++i) {
    int u = read(), k = read();
    query[u].push_back((node){i, k});
  }
  for (int u = 1; u <= n; ++u)
    if (!fa[u]) dfs1(u), f[u] = now, now += h[u], dfs2(u);
  for (int i = 1; i <= q; ++i) printf("%d ", ans[i]);
  return 0;
}

```