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;
int n, m, dem, gt[100005], kq[100005], sz[100005], fa[100005], h[100005],
    sl[100005];
map<string, int> mp;
struct pii {
  int F, S;
};
vector<int> a[100005], vec[100005];
vector<pii> q[100005];
map<int, int> mp1[100005];
void DFS(int u) {
  sz[u] = 1;
  for (auto v : a[u]) {
    h[v] = h[u] + 1;
    DFS(v);
    sz[u] += sz[v];
  }
}
void DSU(int u, int keep) {
  int ma = -1, luu = -1;
  for (auto v : a[u])
    if (ma < sz[v]) {
      ma = sz[v];
      luu = v;
    }
  for (auto v : a[u])
    if (v != luu) DSU(v, 0);
  if (luu != -1) {
    DSU(luu, 1);
    swap(vec[u], vec[luu]);
  }
  vec[u].push_back(u);
  if (mp1[h[u]][gt[u]] == 0) sl[h[u]]++;
  mp1[h[u]][gt[u]]++;
  for (auto v : a[u])
    if (v != luu) {
      for (auto t : vec[v]) {
        mp1[h[t]][gt[t]]++;
        if (mp1[h[t]][gt[t]] == 1) sl[h[t]]++;
        vec[u].push_back(t);
      }
    }
  for (auto x : q[u]) {
    int v = h[u] + x.F, vt = x.S;
    if (v > n)
      kq[vt] = 0;
    else
      kq[vt] = sl[v];
  }
  if (keep == 0) {
    for (auto v : vec[u]) {
      mp1[h[v]][gt[v]]--;
      if (mp1[h[v]][gt[v]] == 0) sl[h[v]]--;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i -= -1) {
    string s;
    int u;
    cin >> s >> u;
    if (mp[s] == 0) {
      gt[i] = ++dem;
      mp[s] = dem;
    } else
      gt[i] = mp[s];
    fa[i] = u;
    a[u].push_back(i);
  }
  for (int i = 1; i <= n; i -= -1)
    if (fa[i] == 0) DFS(i);
  cin >> m;
  for (int i = 1; i <= m; i -= -1) {
    int u, v;
    cin >> u >> v;
    q[u].push_back({v, i});
  }
  for (int i = 1; i <= n; i -= -1)
    if (fa[i] == 0) DSU(i, 0);
  for (int i = 1; i <= m; i -= -1) cout << kq[i] << '\n';
}