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 = 1000 * 100 * 2;
vector<int> g[maxn], q[maxn], qi[maxn], rt;
char S[100000];
string s[maxn], ts[maxn];
int n, m, nm[maxn], ans[maxn], l[maxn], zz;
vector<set<int> > T[maxn];
void mer(set<int> &l, set<int> &r) {
  if (l.size() < r.size()) l.swap(r);
  for (set<int>::iterator it = r.begin(); it != r.end(); it++) l.insert(*it);
}
int nh;
void dfs(int v, int h = 0) {
  nh++;
  for (int i = 0; i < q[v].size(); i++) {
    int th = h + q[v][i];
    l[qi[v][i]] = T[th].size();
  }
  set<int> tst;
  tst.insert(lower_bound(ts, ts + zz, s[v]) - ts);
  T[h].push_back(tst);
  for (int i = 0; i < g[v].size(); i++) dfs(g[v][i], h + 1);
  for (int i = 0; i < q[v].size(); i++) {
    int th = h + q[v][i], tl = l[qi[v][i]], tr = T[th].size();
    if (tl == tr) continue;
    for (int j = tr - 1; j - 1 >= tl; j--) {
      mer(T[th][j - 1], T[th][j]);
      T[th].pop_back();
    }
    ans[qi[v][i]] = T[th][tl].size();
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int p;
    scanf("%s %d", S, &p);
    s[i] = ts[i] = string(S, S + strlen(S));
    p--;
    if (p != -1)
      g[p].push_back(i);
    else
      rt.push_back(i);
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int v, k;
    scanf("%d %d", &v, &k);
    v--;
    q[v].push_back(k);
    qi[v].push_back(i);
  }
  sort(ts, ts + n);
  zz = unique(ts, ts + n) - ts;
  for (int i = 0; i < rt.size(); i++) dfs(rt[i], 0);
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
  return 0;
}
