#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
struct Edge {
  int to, next;
} e[MAXN << 1];
struct Query {
  int dep, next, id;
} q[MAXN];
int n, root, dep[MAXN], ans[MAXN], mx[MAXN], fa[19][MAXN], head[MAXN],
    qhead[MAXN], cnt, qcnt, m;
int mem[MAXN * 21], top[MAXN], val[MAXN];
int *f[MAXN], *now = mem + 3;
inline void insert(int u, int v) {
  e[++cnt].to = v, e[cnt].next = head[u], head[u] = cnt;
}
inline void qinsert(int u, int val, int id) {
  q[++qcnt].dep = val, q[qcnt].next = qhead[u], qhead[u] = qcnt,
  q[qcnt].id = id;
}
void dfs(int x) {
  mx[x] = x;
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    fa[0][v] = x;
    dep[v] = dep[x] + 1;
    dfs(v);
    if (dep[mx[v]] > dep[mx[x]]) mx[x] = mx[v];
  }
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    if (mx[v] != mx[x] || x == 0) {
      f[mx[v]] = (now += (dep[mx[v]] - dep[x] + 1));
      now += 2;
    }
  }
}
void get(int x) {
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    get(v);
    if (mx[v] == mx[x]) f[x] = f[v] - 1;
  }
  f[x][0] = 1;
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    if (mx[v] == mx[x]) continue;
    for (int j = 0; j <= dep[mx[v]] - dep[x]; ++j) f[x][j + 1] += f[v][j];
  }
  for (int i = qhead[x]; i; i = q[i].next) {
    int v = q[i].dep;
    ans[q[i].id] = f[x][v] - 1;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int f;
    scanf("%d", &f);
    insert(f, i);
  }
  dfs(root);
  for (int j = 1; j <= 18; ++j)
    for (int i = 1; i <= n; ++i) fa[j][i] = fa[j - 1][fa[j - 1][i]];
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    int o = x;
    for (int i = 0; i <= 18; ++i)
      if (y & (1 << i)) o = fa[i][o];
    if (!o) {
      ans[i] = 0;
      continue;
    }
    qinsert(o, y, i);
  }
  get(root);
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
  return 0;
}

### 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 = 100005;
struct Edge {
  int to, next;
} e[MAXN << 1];
struct Query {
  int dep, next, id;
} q[MAXN];
int n, root, dep[MAXN], ans[MAXN], mx[MAXN], fa[19][MAXN], head[MAXN],
    qhead[MAXN], cnt, qcnt, m;
int mem[MAXN * 21], top[MAXN], val[MAXN];
int *f[MAXN], *now = mem + 3;
inline void insert(int u, int v) {
  e[++cnt].to = v, e[cnt].next = head[u], head[u] = cnt;
}
inline void qinsert(int u, int val, int id) {
  q[++qcnt].dep = val, q[qcnt].next = qhead[u], qhead[u] = qcnt,
  q[qcnt].id = id;
}
void dfs(int x) {
  mx[x] = x;
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    fa[0][v] = x;
    dep[v] = dep[x] + 1;
    dfs(v);
    if (dep[mx[v]] > dep[mx[x]]) mx[x] = mx[v];
  }
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    if (mx[v] != mx[x] || x == 0) {
      f[mx[v]] = (now += (dep[mx[v]] - dep[x] + 1));
      now += 2;
    }
  }
}
void get(int x) {
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    get(v);
    if (mx[v] == mx[x]) f[x] = f[v] - 1;
  }
  f[x][0] = 1;
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    if (mx[v] == mx[x]) continue;
    for (int j = 0; j <= dep[mx[v]] - dep[x]; ++j) f[x][j + 1] += f[v][j];
  }
  for (int i = qhead[x]; i; i = q[i].next) {
    int v = q[i].dep;
    ans[q[i].id] = f[x][v] - 1;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int f;
    scanf("%d", &f);
    insert(f, i);
  }
  dfs(root);
  for (int j = 1; j <= 18; ++j)
    for (int i = 1; i <= n; ++i) fa[j][i] = fa[j - 1][fa[j - 1][i]];
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    int o = x;
    for (int i = 0; i <= 18; ++i)
      if (y & (1 << i)) o = fa[i][o];
    if (!o) {
      ans[i] = 0;
      continue;
    }
    qinsert(o, y, i);
  }
  get(root);
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
  return 0;
}

```