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;
long long power(long long x, long long y) {
  long long res = 1LL;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long inv(long long n) { return power(n, 1000000007 - 2) % 1000000007; }
long long isprime(long long n) {
  if (n < 2) return 0;
  long long i;
  for (i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
void files() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
const long long N = 2e5 + 10;
vector<long long> a[200005];
long long par[200005][30], dep[200005], sz[200005], ans[200005], big[200005];
set<string> cnt[200005];
string nms[200005];
vector<pair<long long, long long> > qrs[200005];
void dfs1(long long v, long long p, long long d) {
  if (v) par[v][0] = p;
  dep[v] = d;
  sz[v] = 1;
  for (auto x : a[v]) {
    if (x == p) continue;
    dfs1(x, v, d + 1);
    sz[v] += sz[x];
  }
}
void add(long long v, long long par, long long val) {
  if (val == -1)
    cnt[dep[v]].erase(nms[v]);
  else
    cnt[dep[v]].insert(nms[v]);
  for (auto x : a[v]) {
    if (x == par || big[x]) continue;
    add(x, v, val);
  }
}
void dfs2(long long v, long long par, long long keep) {
  long long mx = 0;
  long long bigc = 0;
  for (auto x : a[v]) {
    if (x == par) continue;
    if (sz[x] > mx) {
      mx = sz[x];
      bigc = x;
    }
  }
  for (auto x : a[v]) {
    if (x == par || x == bigc) continue;
    dfs2(x, v, 0);
  }
  if (bigc) dfs2(bigc, v, 1);
  big[bigc] = 1;
  add(v, par, 1);
  for (auto [d, id] : qrs[v]) {
    ans[id] = cnt[d].size();
  }
  big[bigc] = 0;
  if (keep == 0) add(v, par, -1);
}
void solve() {
  long long n;
  cin >> n;
  long long i, x, j;
  for (i = 1; i <= n; i++) {
    cin >> nms[i] >> x;
    a[x].push_back(i);
    a[i].push_back(x);
  }
  dfs1(0, 0, 0);
  long long q;
  cin >> q;
  long long id = 1;
  while (q--) {
    long long dn;
    cin >> x >> dn;
    qrs[x].push_back({dep[x] + dn, id});
    id++;
  }
  dfs2(0, 0, 0);
  for (i = 1; i < id; i++) cout << ans[i] << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}
