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
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100123, lgn = 19;
int f[lgn][maxn], sz[maxn], big[maxn], ans[maxn], cnt[maxn], depth[maxn];
vector<vector<int> > G;
vector<vector<pair<int, int> > > Q;
vector<int> roots;
void dfs_sz(int u = 0, int p = -1, int h = 0) {
  sz[u] = 1;
  depth[u] = h;
  f[0][u] = p;
  for (int i = 1; i < lgn; ++i) f[i][u] = f[i - 1][f[i - 1][u]];
  for (auto v : G[u]) {
    if (v != p) dfs_sz(v, u, h + 1), sz[u] += sz[v];
  }
}
int kth_ancestor(int v, int k) {
  for (int i = lgn - 1; i >= 0; --i) {
    if (k >= (1 << i)) v = f[i][v], k -= 1 << i;
  }
  return v;
}
void add(int u, int p, int h, int x) {
  cnt[h] += x;
  for (auto v : G[u]) {
    if (v != p && !big[v]) add(v, u, h + 1, x);
  }
}
void dfs(int u = 0, int p = -1, int h = 0, bool keep = false) {
  int mx = -1, bigChild = -1;
  for (auto v : G[u]) {
    if (v != p && sz[v] > mx) mx = sz[v], bigChild = v;
  }
  for (auto v : G[u]) {
    if (v != p && v != bigChild) dfs(v, u, h + 1, 0);
  }
  if (bigChild != -1) dfs(bigChild, u, h + 1, 1), big[bigChild] = 1;
  add(u, p, h, 1);
  for (auto q : Q[u]) {
    int v, i;
    tie(v, i) = q;
    ans[i] = cnt[h + v] - 1;
  }
  if (bigChild != -1) big[bigChild] = 0;
  if (!keep) add(u, p, h, -1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  G.resize(n);
  Q.resize(n);
  for (int i = 0; i < n; ++i) {
    int p;
    cin >> p;
    if (p)
      G[p - 1].push_back(i);
    else
      roots.push_back(i);
  }
  int q;
  cin >> q;
  for (auto root : roots) dfs_sz(root, root);
  for (int i = 0; i < q; ++i) {
    int v, p;
    cin >> v >> p;
    --v;
    if (depth[v] < p) continue;
    v = kth_ancestor(v, p);
    Q[v].emplace_back(p, i);
  }
  for (auto root : roots) dfs(root);
  for (int i = 0; i < q; ++i) cout << ans[i] << " ";
}
