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;
const int mod = 1000000007;
const int N = 200100;
const int INF = 0x3f3f3f3f;
void MOD(long long &a) {
  if (a >= mod) a -= mod;
}
void MOD(long long &a, long long c) {
  if (a >= c) a -= c;
}
void ADD(long long &a, long long b) {
  a += b;
  MOD(a);
}
void ADD(long long &a, long long b, long long c) {
  a += b;
  MOD(a, c);
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}
long long qpow(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c;
    b /= 2;
  }
  return ans;
}
int n, head[N], tot = 1, sz[N], son[N], m, fa[N], depth[N], val[N], tid = 1,
                num[N];
bool skip[N];
pair<int, int> ques[N];
map<int, int> ans[N];
map<int, int> cnt[N];
map<string, int> id;
struct EDGE {
  int to, nex;
} e[N * 2];
void addedge(int u, int v) {
  e[tot].to = v;
  e[tot].nex = head[u];
  head[u] = tot++;
}
void dfssz(int now, int fa, int dep) {
  sz[now]++;
  depth[now] = dep;
  for (int i = head[now]; i != -1; i = e[i].nex) {
    if (e[i].to == fa) continue;
    dfssz(e[i].to, now, dep + 1);
    sz[now] += sz[e[i].to];
    if (sz[son[now]] < sz[e[i].to]) son[now] = e[i].to;
  }
}
void add(int now, int fa, int k) {
  cnt[depth[now]][val[now]] += k;
  if (k > 0 && cnt[depth[now]][val[now]] == 1) num[depth[now]]++;
  if (k < 0 && cnt[depth[now]][val[now]] == 0) num[depth[now]]--;
  for (int i = head[now]; i != -1; i = e[i].nex)
    if (e[i].to != fa && !skip[e[i].to]) add(e[i].to, now, k);
}
void dfs(int now, int fa, bool keep) {
  for (int i = head[now]; i != -1; i = e[i].nex)
    if (e[i].to != fa && e[i].to != son[now]) dfs(e[i].to, now, 0);
  if (son[now]) dfs(son[now], now, 1), skip[son[now]] = 1;
  add(now, fa, 1);
  if (((int)ans[now].size())) {
    for (auto it = ans[now].begin(); it != ans[now].end(); it++) {
      ans[now][(*it).first] = num[(*it).first + depth[now]];
    }
  }
  if (son[now]) skip[son[now]] = 0;
  if (!keep) add(now, fa, -1);
}
int main() {
  ios::sync_with_stdio(false);
  memset(head, -1, sizeof(head));
  cin >> n;
  for (int i = 1, u; i <= n; i++) {
    string s;
    cin >> s >> u;
    fa[i] = u;
    if (u) {
      addedge(i, u);
      addedge(u, i);
    }
    if (!id[s]) id[s] = tid++;
    val[i] = id[s];
  }
  cin >> m;
  for (int i = 1, u, v; i <= m; i++) {
    cin >> ques[i].first >> ques[i].second;
    ans[ques[i].first][ques[i].second] = 0;
  }
  for (int i = 1; i <= n; i++)
    if (!fa[i]) dfssz(i, i, 1);
  for (int i = 1; i <= n; i++)
    if (!fa[i]) dfs(i, i, 0);
  for (int i = 1; i <= m; i++)
    cout << ans[ques[i].first][ques[i].second] << endl;
}