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 Head[500010], Ver[500010], Next[500010];
long long root[500010], Cntroot;
long long col[500010];
long long tot = 0;
long long ans = 0;
map<string, long long> ss[500010];
long long Ans[500010];
long long Vis[500010];
string s[500010];
long long fa[500010][25], dep[500010], siz[500010], Son[500010];
vector<pair<long long, long long> > Que[500010];
long long cnt[500010 >> 1];
void Add(long long x, long long y) {
  Ver[++tot] = y;
  Next[tot] = Head[x];
  Head[x] = tot;
}
long long n, m;
void dfs1(long long x, long long Fa) {
  siz[x] = 1;
  dep[x] = dep[Fa] + 1;
  fa[x][0] = Fa;
  for (long long i = 1; i <= 20; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  long long Mx = -1;
  for (long long i = Head[x]; i; i = Next[i]) {
    long long y = Ver[i];
    if (y == Fa) continue;
    dfs1(y, x);
    siz[x] += siz[y];
    if (siz[y] > Mx) Mx = siz[y], Son[x] = y;
  }
}
void Calc(long long x, long long Fa, long long opt) {
  if (!ss[dep[x]][s[x]]) cnt[dep[x]]++;
  ss[dep[x]][s[x]] += opt;
  if (!ss[dep[x]][s[x]]) cnt[dep[x]]--;
  for (long long i = Head[x]; i; i = Next[i]) {
    long long y = Ver[i];
    if (y == Fa || Vis[y]) continue;
    Calc(y, x, opt);
  }
}
void dfs2(long long x, long long Fa, long long opt) {
  for (long long i = Head[x]; i; i = Next[i]) {
    long long y = Ver[i];
    if (y == Fa || y == Son[x]) continue;
    dfs2(y, x, 0);
  }
  if (Son[x]) {
    dfs2(Son[x], x, 1);
    Vis[Son[x]] = 1;
  }
  Calc(x, Fa, 1);
  for (long long i = 0; i < Que[x].size(); i++) {
    long long id = Que[x][i].first, d = Que[x][i].second;
    Ans[id] = cnt[d];
  }
  if (Son[x]) Vis[Son[x]] = 0;
  if (!opt) Calc(x, Fa, -1);
}
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> s[i];
    scanf("%lld", &x);
    if (!x)
      root[++Cntroot] = i;
    else
      Add(x, i), Add(i, x);
  }
  for (long long i = 1; i <= Cntroot; i++) dfs1(root[i], 0);
  scanf("%lld", &m);
  for (long long i = 1; i <= m; i++) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    Que[x].push_back(make_pair(i, dep[x] + y));
  }
  for (long long i = 1; i <= Cntroot; i++) dfs2(root[i], 0, 0);
  for (long long i = 1; i <= m; i++) printf("%lld\n", Ans[i]);
  return 0;
}
