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 = 5e5 + 100;
vector<int> g[MAXN], Q[MAXN];
int sz[MAXN], c[MAXN], big[MAXN], ans[MAXN];
int v[MAXN], ht[MAXN], depth[MAXN];
map<string, int> cnt[MAXN];
map<int, int> res[MAXN];
string a[MAXN];
map<string, int>::iterator it;
void dfs0(int x, int h) {
  sz[x] = 1;
  depth[x] = h;
  for (int i = 0; i < g[x].size(); ++i) {
    int u = g[x][i];
    dfs0(u, h + 1);
    sz[x] += sz[u];
  }
}
void add(int x) {
  cnt[depth[x]][a[x]]++;
  for (int i = 0; i < g[x].size(); ++i) {
    int u = g[x][i];
    if (!big[u]) add(u);
  }
}
void rem(int v) {
  it = cnt[depth[v]].find(a[v]);
  it->second--;
  if (!it->second) cnt[depth[v]].erase(it);
  for (int i = 0; i < g[v].size(); ++i) {
    int u = g[v][i];
    if (!big[u]) rem(u);
  }
}
void dfs(int x, int keep) {
  int mx = 0, bigc = 0;
  for (int i = 0; i < g[x].size(); ++i) {
    int u = g[x][i];
    if (mx < sz[u]) mx = sz[u], bigc = u;
  }
  for (int i = 0; i < g[x].size(); ++i) {
    int u = g[x][i];
    if (u != bigc) dfs(u, 1);
  }
  if (bigc) dfs(bigc, 0), big[bigc] = 1;
  add(x);
  for (int i = 0; i < Q[x].size(); ++i) {
    int h = Q[x][i];
    res[x][h + depth[x]] = cnt[h + depth[x]].size();
  }
  if (bigc) big[bigc] = 0;
  if (keep) rem(x);
}
int main() {
  ios_base::sync_with_stdio(0);
  ;
  int n, q;
  cin >> n;
  int x, y;
  a[0] = "hell";
  for (int i = 1; i < n + 1; ++i) {
    cin >> a[i] >> x;
    g[x].push_back(i);
  }
  dfs0(0, 0);
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> v[i] >> ht[i];
    Q[v[i]].push_back(ht[i]);
  }
  dfs(0, 0);
  for (int i = 0; i < q; ++i) {
    cout << res[v[i]][ht[i] + depth[v[i]]] << endl;
  }
}
