#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000033, LN = 20;
int pa[LN][MAXN], st[MAXN], en[MAXN], dep[MAXN], pool;
vector<int> g[MAXN], at[MAXN];
void dfs(int u, int d) {
  dep[u] = d;
  st[u] = ++pool;
  at[d].push_back(pool);
  for (auto v : g[u]) dfs(v, d + 1);
  en[u] = pool;
}
int kthAncestor(int u, int k) {
  for (int i = LN - 1; i >= 0; --i)
    if (k & (1 << i)) u = pa[i][u];
  return u;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &pa[0][i]), g[pa[0][i]].push_back(i);
  for (int i = 1; i <= n; ++i)
    if (!st[i]) dfs(i, 1);
  for (int j = 1; j < LN; ++j)
    for (int i = 1; i <= n; ++i) pa[j][i] = pa[j - 1][pa[j - 1][i]];
  int q;
  scanf("%d", &q);
  while (q--) {
    int v, k;
    scanf("%d %d", &v, &k);
    int p = kthAncestor(v, k);
    int l = st[p], r = en[p], d = dep[v];
    if (p == 0)
      printf("0 ");
    else
      printf("%lu ", upper_bound(at[d].begin(), at[d].end(), r) -
                         lower_bound(at[d].begin(), at[d].end(), l) - 1);
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 1000033, LN = 20;
int pa[LN][MAXN], st[MAXN], en[MAXN], dep[MAXN], pool;
vector<int> g[MAXN], at[MAXN];
void dfs(int u, int d) {
  dep[u] = d;
  st[u] = ++pool;
  at[d].push_back(pool);
  for (auto v : g[u]) dfs(v, d + 1);
  en[u] = pool;
}
int kthAncestor(int u, int k) {
  for (int i = LN - 1; i >= 0; --i)
    if (k & (1 << i)) u = pa[i][u];
  return u;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &pa[0][i]), g[pa[0][i]].push_back(i);
  for (int i = 1; i <= n; ++i)
    if (!st[i]) dfs(i, 1);
  for (int j = 1; j < LN; ++j)
    for (int i = 1; i <= n; ++i) pa[j][i] = pa[j - 1][pa[j - 1][i]];
  int q;
  scanf("%d", &q);
  while (q--) {
    int v, k;
    scanf("%d %d", &v, &k);
    int p = kthAncestor(v, k);
    int l = st[p], r = en[p], d = dep[v];
    if (p == 0)
      printf("0 ");
    else
      printf("%lu ", upper_bound(at[d].begin(), at[d].end(), r) -
                         lower_bound(at[d].begin(), at[d].end(), l) - 1);
  }
}

```