Problem: Polycarpus got hold of a family tree. The found tree describes the family relations of n people, numbered from 1 to n. Every person in this tree has at most one direct ancestor. Also, each person in the tree has a name, the names are not necessarily unique.

We call the man with a number a a 1-ancestor of the man with a number b, if the man with a number a is a direct ancestor of the man with a number b.

We call the man with a number a a k-ancestor (k > 1) of the man with a number b, if the man with a number b has a 1-ancestor, and the man with a number a is a (k - 1)-ancestor of the 1-ancestor of the man with a number b.

In the tree the family ties do not form cycles. In other words there isn't a person who is his own direct or indirect ancestor (that is, who is an x-ancestor of himself, for some x, x > 0).

We call a man with a number a the k-son of the man with a number b, if the man with a number b is a k-ancestor of the man with a number a.

Polycarpus is very much interested in how many sons and which sons each person has. He took a piece of paper and wrote m pairs of numbers vi, ki. Help him to learn for each pair vi, ki the number of distinct names among all names of the ki-sons of the man with number vi.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. Next n lines contain the description of people in the tree. The i-th line contains space-separated string si and integer ri (0 ≤ ri ≤ n), where si is the name of the man with a number i, and ri is either the number of the direct ancestor of the man with a number i or 0, if the man with a number i has no direct ancestor. 

The next line contains a single integer m (1 ≤ m ≤ 105) — the number of Polycarpus's records. Next m lines contain space-separated pairs of integers. The i-th line contains integers vi, ki (1 ≤ vi, ki ≤ n).

It is guaranteed that the family relationships do not form cycles. The names of all people are non-empty strings, consisting of no more than 20 lowercase English letters.

Output

Print m whitespace-separated integers — the answers to Polycarpus's records. Print the answers to the records in the order, in which the records occur in the input.

Examples

Input

6
pasha 0
gerald 1
gerald 1
valera 2
igor 3
olesya 1
5
1 1
1 2
1 3
3 1
6 1


Output

2
2
0
1
0


Input

6
valera 0
valera 1
valera 1
gerald 0
valera 4
kolya 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

1
0
0
0
2
0
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
mt19937_64 mrand(chrono::steady_clock::now().time_since_epoch().count());
vector<int> g[100005], rt;
vector<pair<int, int>> q[100005];
int sz[100005], son[100005], dep[100005], f[100005][18], ans[100005];
void dfs(int u, int fa) {
  sz[u] = 1;
  for (int v : g[u]) {
    if (v == fa) continue;
    dep[v] = dep[u] + 1;
    f[v][0] = u;
    for (int i = 1; i <= 17; ++i) f[v][i] = f[f[v][i - 1]][i - 1];
    dfs(v, u);
    sz[u] += sz[v];
    if (sz[son[u]] < sz[v]) son[u] = v;
  }
}
int cnt[200005];
void add(int u, int fa) {
  cnt[dep[u]]++;
  for (int v : g[u]) {
    if (v == fa) continue;
    add(v, u);
  }
}
void del(int u, int fa) {
  cnt[dep[u]]--;
  for (int v : g[u]) {
    if (v == fa) continue;
    del(v, u);
  }
}
void dfs2(int u, int fa, int keep) {
  for (int v : g[u]) {
    if (v == son[u] || v == fa) continue;
    dfs2(v, u, 0);
  }
  if (son[u]) dfs2(son[u], u, 1);
  cnt[dep[u]]++;
  for (int v : g[u])
    if (v != son[u] && v != fa) add(v, u);
  for (auto p : q[u]) ans[p.second] = cnt[p.first + dep[u]];
  if (!keep) {
    cnt[dep[u]]--;
    for (int v : g[u])
      if (v != fa) del(v, u);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    if (x)
      g[x].push_back(i);
    else
      rt.push_back(i);
  }
  for (int i : rt) dfs(i, 0);
  int t;
  cin >> t;
  for (int i = 1; i <= t; ++i) {
    int v, p;
    cin >> v >> p;
    for (int j = 17; j >= 0; --j)
      if (p >> j & 1) v = f[v][j];
    if (!v)
      ans[i] = 1;
    else
      q[v].push_back({p, i});
  }
  for (int i : rt) dfs2(i, 0, 0);
  for (int i = 1; i <= t; ++i) cout << ans[i] - 1 << " ";
  cout << endl;
}