#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
long long add(long long a, long long b) {
  long long ret = a + b;
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
long long subtract(long long a, long long b) {
  long long ret = a - b;
  if (ret < 0) ret += 1000000007;
  return ret;
}
long long mult(long long a, long long b) {
  long long ret = a * b;
  if (ret >= 1000000007) ret %= 1000000007;
  return ret;
}
long long bigmod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = mult(ret, a);
    b >>= 1;
    a = mult(a, a);
  }
  return ret;
}
long long inverse(long long n) { return bigmod(n, 1000000007 - 2); }
bool base[1000010];
vector<int> primes;
void sieve(int mx) {
  mx += 10;
  int x = sqrt(mx);
  for (int i = 3; i <= x; i += 2)
    if (base[i] == 0)
      for (int j = i * i, k = i << 1; j < mx; j += k) base[j] = 1;
  primes.push_back(2);
  for (int i = 3; i < mx; i += 2)
    if (base[i] == 0) primes.push_back(i);
}
bool checkBit(int n, int i) { return (n & (1 << i)); }
int setBit(int n, int i) { return (n | (1 << i)); }
int resetBit(int n, int i) { return (n & (~(1 << i))); }
int sub[200010], parent[200010][22], cnt[200010], big[200010], ans[200010];
vector<int> e[200010];
vector<pair<int, int> > q[200010];
void dfs(int u) {
  sub[u] = 1;
  for (int &v : e[u]) {
    dfs(v);
    sub[u] += sub[v];
  }
}
void add(int u, int d, int x) {
  cnt[d] += x;
  for (int &v : e[u])
    if (!big[v]) {
      add(v, d + 1, x);
    }
}
void dfs(int u, int d, int keep) {
  int mx = -1, bigchild = -1;
  for (int &v : e[u])
    if (sub[v] > mx) {
      mx = sub[v], bigchild = v;
    }
  for (int &v : e[u])
    if (v != bigchild) {
      dfs(v, d + 1, 0);
    }
  if (bigchild != -1) {
    dfs(bigchild, d + 1, 1);
    big[bigchild] = 1;
  }
  add(u, d, 1);
  for (pair<int, int> &it : q[u]) ans[it.second] = cnt[it.first + d] - 1;
  if (bigchild + 1) {
    big[bigchild] = 0;
  }
  if (!keep) {
    add(u, d, -1);
  }
}
int main() {
  int n, Q, u, h;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &parent[i][0]);
    if (parent[i][0]) {
      e[parent[i][0]].push_back(i);
    } else
      parent[i][0] = -1;
  }
  for (int i = 1; i <= n; ++i)
    if (parent[i][0] == -1) {
      dfs(i);
    }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= 20; ++j) parent[i][j] = -1;
  for (int j = 1; (1 << j) <= n; ++j)
    for (int i = 1; i <= n; ++i) {
      if (parent[i][j - 1] != -1)
        parent[i][j] = parent[parent[i][j - 1]][j - 1];
    }
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    scanf("%d%d", &u, &h);
    for (int j = 19; j >= 0; --j)
      if (checkBit(h, j)) {
        u = parent[u][j];
      }
    if (u + 1) q[u].push_back({h, i});
  }
  for (int i = 1; i <= n; ++i)
    if (parent[i][0] == -1) {
      dfs(i, 0, 0);
    }
  for (int i = 0; i < Q; ++i) printf("%d ", ans[i]);
  puts("");
  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 double PI = acos(-1);
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
long long add(long long a, long long b) {
  long long ret = a + b;
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
long long subtract(long long a, long long b) {
  long long ret = a - b;
  if (ret < 0) ret += 1000000007;
  return ret;
}
long long mult(long long a, long long b) {
  long long ret = a * b;
  if (ret >= 1000000007) ret %= 1000000007;
  return ret;
}
long long bigmod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = mult(ret, a);
    b >>= 1;
    a = mult(a, a);
  }
  return ret;
}
long long inverse(long long n) { return bigmod(n, 1000000007 - 2); }
bool base[1000010];
vector<int> primes;
void sieve(int mx) {
  mx += 10;
  int x = sqrt(mx);
  for (int i = 3; i <= x; i += 2)
    if (base[i] == 0)
      for (int j = i * i, k = i << 1; j < mx; j += k) base[j] = 1;
  primes.push_back(2);
  for (int i = 3; i < mx; i += 2)
    if (base[i] == 0) primes.push_back(i);
}
bool checkBit(int n, int i) { return (n & (1 << i)); }
int setBit(int n, int i) { return (n | (1 << i)); }
int resetBit(int n, int i) { return (n & (~(1 << i))); }
int sub[200010], parent[200010][22], cnt[200010], big[200010], ans[200010];
vector<int> e[200010];
vector<pair<int, int> > q[200010];
void dfs(int u) {
  sub[u] = 1;
  for (int &v : e[u]) {
    dfs(v);
    sub[u] += sub[v];
  }
}
void add(int u, int d, int x) {
  cnt[d] += x;
  for (int &v : e[u])
    if (!big[v]) {
      add(v, d + 1, x);
    }
}
void dfs(int u, int d, int keep) {
  int mx = -1, bigchild = -1;
  for (int &v : e[u])
    if (sub[v] > mx) {
      mx = sub[v], bigchild = v;
    }
  for (int &v : e[u])
    if (v != bigchild) {
      dfs(v, d + 1, 0);
    }
  if (bigchild != -1) {
    dfs(bigchild, d + 1, 1);
    big[bigchild] = 1;
  }
  add(u, d, 1);
  for (pair<int, int> &it : q[u]) ans[it.second] = cnt[it.first + d] - 1;
  if (bigchild + 1) {
    big[bigchild] = 0;
  }
  if (!keep) {
    add(u, d, -1);
  }
}
int main() {
  int n, Q, u, h;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &parent[i][0]);
    if (parent[i][0]) {
      e[parent[i][0]].push_back(i);
    } else
      parent[i][0] = -1;
  }
  for (int i = 1; i <= n; ++i)
    if (parent[i][0] == -1) {
      dfs(i);
    }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= 20; ++j) parent[i][j] = -1;
  for (int j = 1; (1 << j) <= n; ++j)
    for (int i = 1; i <= n; ++i) {
      if (parent[i][j - 1] != -1)
        parent[i][j] = parent[parent[i][j - 1]][j - 1];
    }
  scanf("%d", &Q);
  for (int i = 0; i < Q; ++i) {
    scanf("%d%d", &u, &h);
    for (int j = 19; j >= 0; --j)
      if (checkBit(h, j)) {
        u = parent[u][j];
      }
    if (u + 1) q[u].push_back({h, i});
  }
  for (int i = 1; i <= n; ++i)
    if (parent[i][0] == -1) {
      dfs(i, 0, 0);
    }
  for (int i = 0; i < Q; ++i) printf("%d ", ans[i]);
  puts("");
  return 0;
}

```