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;
struct pii {
  int x, y;
};
inline int lowbit(const int x) { return x & (-x); };
const int N = 100005;
struct Tree {
  vector<int> adj[N];
  int ld[N], rd[N], sz[N], id[N];
  int big[N], dep[N];
  int n, ind, rt;
  void trprint() {
    cout << "rt=" << rt << ' ' << "n=" << n << endl;
    for (int i = 1; i <= n; ++i)
      cout << "i=" << i << ' ' << ld[i] << ' ' << rd[i] << ' ' << sz[i] << ' '
           << big[i] << endl;
  }
  void trinit() {
    for (int i = 1; i <= n; ++i) adj[i].clear();
    n = ind = 0;
    rt = 1;
  }
  inline void addedge(const int x, const int y) {
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  void trdfs(int x, int pre) {
    ld[x] = ++ind;
    id[ind] = x;
    sz[x] = 1;
    big[x] = 0;
    dep[x] = dep[pre] + 1;
    for (auto &y : adj[x])
      if (y != pre) {
        trdfs(y, x);
        sz[x] += sz[y];
        if (sz[y] > sz[big[x]]) big[x] = y;
      }
    rd[x] = ld[x] + sz[x] - 1;
  }
};
pii q[N];
vector<int> query[N];
int cnt[N];
int ans[N];
string nam[N];
map<string, int> mp[N];
struct DsuOnTree : Tree {
  void dsuprint() {
    for (int i = 1; i <= n; ++i)
      cout << "i=" << i << ' ' << nam[i] << ' ' << ans[i] << endl;
  }
  inline void add(const int x, const int t) {
    int nw = (mp[dep[x]][nam[x]] += t);
    if (nw == 1 && t == 1) ++cnt[dep[x]];
    if (nw == 0 && t == -1) --cnt[dep[x]];
  }
  void calcans(const int u) {
    for (const auto &i : query[u])
      if (q[i].y <= n) ans[i] = cnt[q[i].y];
  }
  void relax(const int x) {
    for (int i = ld[x]; i <= rd[x]; ++i) add(id[i], -1);
  }
  void dfs(int x, int pre, int keep) {
    for (auto &y : adj[x])
      if (y != pre && y != big[x]) dfs(y, x, 0);
    if (big[x]) dfs(big[x], x, 1);
    add(x, 1);
    for (auto &y : adj[x])
      if (y != pre && y != big[x])
        for (int i = ld[y]; i <= rd[y]; ++i) add(id[i], 1);
    calcans(x);
    if (!keep) relax(x);
  }
} tr;
int n, m;
int main() {
  tr.trinit();
  cin >> n;
  ++n;
  tr.n = n;
  for (int i = 2, x, y; i <= n; ++i)
    cin >> nam[i], scanf("%d", &x), tr.addedge(x + 1, i);
  tr.trdfs(1, 0);
  cin >> m;
  for (int i = 1, x, y; i <= m; ++i) {
    scanf("%d%d", &x, &y);
    ++x;
    y += tr.dep[x];
    q[i] = {x, y};
    query[x].push_back(i);
  }
  tr.dfs(1, 0, 0);
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}