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 block = 1000;
struct node {
  int idx, l, r, val;
  node() {}
  node(int ii, int ll, int rr, int v) {
    idx = ii;
    l = ll;
    r = rr;
    val = v;
  }
  inline pair<int, int> toPair() const {
    return make_pair(l / block, ((l / block) & 1) ? -r : +r);
  }
};
inline bool operator<(const node &a, const node &b) {
  return a.toPair() < b.toPair();
}
const int N = 3e5 + 100;
int t = 0;
vector<node> qry;
vector<int> ord(N), adj[N], dep(N), in(N, -1), out(N), ans(N);
int up[N][20];
void dfs(int src, int par) {
  in[src] = ++t;
  ord[in[src]] = dep[src];
  for (int i : adj[src]) {
    if (i == par) continue;
    dep[i] = 1 + dep[src];
    up[i][0] = src;
    for (int j = 1; j < 20; j++) up[i][j] = up[up[i][j - 1]][j - 1];
    dfs(i, src);
  }
  out[src] = t;
}
void lift(int &u, int &p) {
  for (int i = 19; i >= 0; i--)
    if ((1 << i) <= p) {
      p -= (1 << i);
      u = up[u][i];
    }
}
int hm[N];
void add(int x) { hm[ord[x]]++; }
void remove(int x) { hm[ord[x]]--; }
void mos() {
  int cur_l = 0, cur_r = -1;
  sort(qry.begin(), qry.end());
  for (auto &q : qry) {
    while (cur_l > q.l) {
      cur_l--;
      add(cur_l);
    }
    while (cur_r > q.r) {
      remove(cur_r);
      cur_r--;
    }
    while (cur_l < q.l) {
      remove(cur_l);
      cur_l++;
    }
    while (cur_r < q.r) {
      cur_r++;
      add(cur_r);
    }
    ans[q.idx] = hm[q.val] - 1;
  }
  return;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x == 0) continue;
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (in[i] == -1) dfs(i, i);
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int u, p;
    cin >> u >> p;
    int d = dep[u];
    lift(u, p);
    if (u == 0) {
      ans[i] = 0;
      continue;
    }
    qry.emplace_back(i, in[u], out[u], d);
  }
  mos();
  for (int i = 1; i <= q; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}
