#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
struct Q {
  int dep, idx;
};
vector<Q> q[maxn];
int ans[maxn];
int c[maxn], sz[maxn], dep[maxn], bigson[maxn];
int p[maxn][20];
vector<int> g[maxn];
void dfs1(int x, int fa) {
  p[x][0] = fa;
  dep[x] = dep[fa] + 1;
  for (int j = (1); j <= (18); j++) p[x][j] = p[p[x][j - 1]][j - 1];
  sz[x] = 1;
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i];
    if (v == fa) continue;
    dfs1(v, x);
    sz[x] += sz[v];
    if (sz[v] > sz[bigson[x]]) bigson[x] = v;
  }
}
void calc(int x, int fa, int val) {
  c[dep[x]] += val;
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i];
    if (v == fa) continue;
    calc(v, x, val);
  }
}
void dfs(int x, int fa, bool isdel) {
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i];
    if (v == fa || v == bigson[x]) continue;
    dfs(v, x, 1);
  }
  if (bigson[x]) dfs(bigson[x], x, 0);
  c[dep[x]]++;
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i];
    if (v == fa || v == bigson[x]) continue;
    calc(v, x, 1);
  }
  for (int i = (0); i < (q[x].size()); i++) ans[q[x][i].idx] = c[q[x][i].dep];
  if (isdel) calc(x, fa, -1);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    int d;
    scanf("%d", &d);
    g[d].push_back(i);
    g[i].push_back(d);
  }
  for (int i = (0); i < (g[0].size()); i++) dfs1(g[0][i], 0);
  int m;
  scanf("%d", &m);
  for (int i = (1); i <= (m); i++) {
    int x, k;
    scanf("%d", &x);
    scanf("%d", &k);
    int d = dep[x];
    for (int j = 18; j >= 0; j--) {
      if (k >= (1 << j)) {
        k -= (1 << j);
        x = p[x][j];
      }
    }
    q[x].push_back((Q){d, i});
  }
  for (int i = (0); i < (g[0].size()); i++) dfs(g[0][i], 0, 1);
  for (int i = (1); i <= (m); i++) printf("%d ", max(0, ans[i] - 1));
}

### Prompt

Please formulate 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 maxn = 1e5 + 5;
const int mod = 1e9 + 7;
struct Q {
  int dep, idx;
};
vector<Q> q[maxn];
int ans[maxn];
int c[maxn], sz[maxn], dep[maxn], bigson[maxn];
int p[maxn][20];
vector<int> g[maxn];
void dfs1(int x, int fa) {
  p[x][0] = fa;
  dep[x] = dep[fa] + 1;
  for (int j = (1); j <= (18); j++) p[x][j] = p[p[x][j - 1]][j - 1];
  sz[x] = 1;
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i];
    if (v == fa) continue;
    dfs1(v, x);
    sz[x] += sz[v];
    if (sz[v] > sz[bigson[x]]) bigson[x] = v;
  }
}
void calc(int x, int fa, int val) {
  c[dep[x]] += val;
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i];
    if (v == fa) continue;
    calc(v, x, val);
  }
}
void dfs(int x, int fa, bool isdel) {
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i];
    if (v == fa || v == bigson[x]) continue;
    dfs(v, x, 1);
  }
  if (bigson[x]) dfs(bigson[x], x, 0);
  c[dep[x]]++;
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i];
    if (v == fa || v == bigson[x]) continue;
    calc(v, x, 1);
  }
  for (int i = (0); i < (q[x].size()); i++) ans[q[x][i].idx] = c[q[x][i].dep];
  if (isdel) calc(x, fa, -1);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    int d;
    scanf("%d", &d);
    g[d].push_back(i);
    g[i].push_back(d);
  }
  for (int i = (0); i < (g[0].size()); i++) dfs1(g[0][i], 0);
  int m;
  scanf("%d", &m);
  for (int i = (1); i <= (m); i++) {
    int x, k;
    scanf("%d", &x);
    scanf("%d", &k);
    int d = dep[x];
    for (int j = 18; j >= 0; j--) {
      if (k >= (1 << j)) {
        k -= (1 << j);
        x = p[x][j];
      }
    }
    q[x].push_back((Q){d, i});
  }
  for (int i = (0); i < (g[0].size()); i++) dfs(g[0][i], 0, 1);
  for (int i = (1); i <= (m); i++) printf("%d ", max(0, ans[i] - 1));
}

```