#include <bits/stdc++.h>
using namespace std;
int i, j, m, n, p, k, root[200001], tot, vis[200001], st[200001], ed[200001],
    top, fox[200001], k1;
int fa[200001][21], deep[100001], x, y, ans;
struct Node {
  int l, r, sum;
} Tree[5000001];
struct Bian {
  int ed, before;
} s[1000001];
void add(int x, int y) {
  s[++k1].ed = y;
  s[k1].before = fox[x];
  fox[x] = k1;
}
void update(int ll, int l, int r, int t) {
  if (l == r)
    Tree[t].sum++;
  else {
    if (ll <= ((l + r) >> 1)) {
      if (!Tree[t].l) Tree[t].l = ++tot;
      update(ll, l, ((l + r) >> 1), Tree[t].l);
    } else {
      if (!Tree[t].r) Tree[t].r = ++tot;
      update(ll, ((l + r) >> 1) + 1, r, Tree[t].r);
    }
    Tree[t].sum = Tree[Tree[t].l].sum + Tree[Tree[t].r].sum;
  }
}
void ask(int Q, int ll, int rr, int l, int r, int t) {
  if (ll <= l && r <= rr)
    ans += Q * Tree[t].sum;
  else {
    if (ll <= ((l + r) >> 1)) ask(Q, ll, rr, l, ((l + r) >> 1), Tree[t].l);
    if (rr > ((l + r) >> 1)) ask(Q, ll, rr, ((l + r) >> 1) + 1, r, Tree[t].r);
  }
}
void Get(int x) {
  int i;
  for (i = 1; i <= 20; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
}
int get(int x, int y) {
  for (i = 20; i >= 0; i--)
    if (y >= (1 << i)) x = fa[x][i], y -= (1 << i);
  return x;
}
void dfs(int x) {
  vis[x] = 1;
  st[x] = ++top;
  Get(x);
  int i;
  for (i = fox[x]; i; i = s[i].before)
    if (!vis[s[i].ed]) {
      deep[s[i].ed] = deep[x] + 1;
      fa[s[i].ed][0] = x;
      dfs(s[i].ed);
    }
  ed[x] = top;
}
int main() {
  scanf("%d", &n);
  for (i = 0; i <= n; i++) root[i] = ++tot;
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x > 0) add(x, i), fa[i][0] = x;
  }
  for (i = 1; i <= n; i++)
    if (!fa[i][0]) dfs(i);
  for (i = 1; i <= n; i++) update(st[i], 1, n, root[deep[i]]);
  scanf("%d", &m);
  for (; m--;) {
    scanf("%d%d", &x, &y);
    int lca = get(x, y);
    if (lca == 0) {
      printf("0 ");
      continue;
    }
    ans = 0;
    ask(1, st[lca], ed[lca], 1, n, root[deep[x]]);
    printf("%d ", ans - 1);
  }
}

### Prompt

Please provide a CPP coded solution 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;
int i, j, m, n, p, k, root[200001], tot, vis[200001], st[200001], ed[200001],
    top, fox[200001], k1;
int fa[200001][21], deep[100001], x, y, ans;
struct Node {
  int l, r, sum;
} Tree[5000001];
struct Bian {
  int ed, before;
} s[1000001];
void add(int x, int y) {
  s[++k1].ed = y;
  s[k1].before = fox[x];
  fox[x] = k1;
}
void update(int ll, int l, int r, int t) {
  if (l == r)
    Tree[t].sum++;
  else {
    if (ll <= ((l + r) >> 1)) {
      if (!Tree[t].l) Tree[t].l = ++tot;
      update(ll, l, ((l + r) >> 1), Tree[t].l);
    } else {
      if (!Tree[t].r) Tree[t].r = ++tot;
      update(ll, ((l + r) >> 1) + 1, r, Tree[t].r);
    }
    Tree[t].sum = Tree[Tree[t].l].sum + Tree[Tree[t].r].sum;
  }
}
void ask(int Q, int ll, int rr, int l, int r, int t) {
  if (ll <= l && r <= rr)
    ans += Q * Tree[t].sum;
  else {
    if (ll <= ((l + r) >> 1)) ask(Q, ll, rr, l, ((l + r) >> 1), Tree[t].l);
    if (rr > ((l + r) >> 1)) ask(Q, ll, rr, ((l + r) >> 1) + 1, r, Tree[t].r);
  }
}
void Get(int x) {
  int i;
  for (i = 1; i <= 20; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
}
int get(int x, int y) {
  for (i = 20; i >= 0; i--)
    if (y >= (1 << i)) x = fa[x][i], y -= (1 << i);
  return x;
}
void dfs(int x) {
  vis[x] = 1;
  st[x] = ++top;
  Get(x);
  int i;
  for (i = fox[x]; i; i = s[i].before)
    if (!vis[s[i].ed]) {
      deep[s[i].ed] = deep[x] + 1;
      fa[s[i].ed][0] = x;
      dfs(s[i].ed);
    }
  ed[x] = top;
}
int main() {
  scanf("%d", &n);
  for (i = 0; i <= n; i++) root[i] = ++tot;
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x > 0) add(x, i), fa[i][0] = x;
  }
  for (i = 1; i <= n; i++)
    if (!fa[i][0]) dfs(i);
  for (i = 1; i <= n; i++) update(st[i], 1, n, root[deep[i]]);
  scanf("%d", &m);
  for (; m--;) {
    scanf("%d%d", &x, &y);
    int lca = get(x, y);
    if (lca == 0) {
      printf("0 ");
      continue;
    }
    ans = 0;
    ask(1, st[lca], ed[lca], 1, n, root[deep[x]]);
    printf("%d ", ans - 1);
  }
}

```